python | Official Python client library for kubernetes | REST library

by kubernetes-client Python Version: v26.1.0a1 License: Apache-2.0

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python is a Python library typically used in Web Services, REST applications. python has no bugs, it has build file available, it has a Permissive License and it has high support. However python has 2 vulnerabilities. You can install using 'pip install python' or download it from GitHub, PyPI.

Python client for the kubernetes API.

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python has a highly active ecosystem.
It has 5430 star(s) with 2969 fork(s). There are 136 watchers for this library.
There were 4 major release(s) in the last 6 months.
There are 91 open issues and 1101 have been closed. On average issues are closed in 164 days. There are 19 open pull requests and 0 closed requests.
It has a positive sentiment in the developer community.
The latest version of python is v26.1.0a1

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python is licensed under the Apache-2.0 License. This license is Permissive.
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Top functions reviewed by kandi - BETA

kandi has reviewed python and discovered the below as its top functions. This is intended to give you an instant insight into python implemented functionality, and help decide if they suit your requirements.

List namespaced Custom Object Objects

Delete collection_namespaced_namespaced_namespace_namespaced_namespace_namespace_id

Delete collection namespaced deployment .

delete collection_namespaced_namespaced_job

Delete collection namespace .

Delete collection endpoint slice

Delete collection namespaced lease .

Delete collection_namespaced_namespaced_namespace_autospace_autosheet_autosheet_autosheet_autosheet_autosheet_autosheet_autospace_autosheet_autospace_autosheet_autosheet_autospace_autospace_autospace_autosheet_autosheet_autosheet_autosheet_autospace_autosheet_autoscheosheet_autosheet_autosheet_autosheet_autosheet_autosheet_autospace_autos

Delete collection_namespaced_csi_csi_storage_capacity

Delete collection namespace binding

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Official Python client library for kubernetes

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Community Discussions

Trending Discussions on python

Python/Docker ImportError: cannot import name 'json' from itsdangerous
Why is it faster to compare strings that match than strings that do not?
Why is `np.sum(range(N))` very slow?
Error while downloading the requirements using pip install (setup command: use_2to3 is invalid.)
Repeatedly removing the maximum average subarray
WARNING: Running pip as the 'root' user
How do I calculate square root in Python?
pip-compile raising AssertionError on its logging handler
ImportError: cannot import name 'url' from 'django.conf.urls' after upgrading to Django 4.0
How did print(*a, a.pop(0)) change?

Trending Discussions on python

QUESTION

Python/Docker ImportError: cannot import name 'json' from itsdangerous

Asked 2022-Mar-31 at 12:49

I am trying to get a Flask and Docker application to work but when I try and run it using my docker-compose up command in my Visual Studio terminal, it gives me an ImportError called ImportError: cannot import name 'json' from itsdangerous. I have tried to look for possible solutions to this problem but as of right now there are not many on here or anywhere else. The only two solutions I could find are to change the current installation of MarkupSafe and itsdangerous to a higher version: https://serverfault.com/questions/1094062/from-itsdangerous-import-json-as-json-importerror-cannot-import-name-json-fr and another one on GitHub that tells me to essentially change the MarkUpSafe and itsdangerous installation again https://github.com/aws/aws-sam-cli/issues/3661, I have also tried to make a virtual environment named veganetworkscriptenv to install the packages but that has also failed as well. I am currently using Flask 2.0.0 and Docker 5.0.0 and the error occurs on line eight in vegamain.py.

Here is the full ImportError that I get when I try and run the program:

veganetworkscript-backend-1  | Traceback (most recent call last):
veganetworkscript-backend-1  |   File "/app/vegamain.py", line 8, in <module>
veganetworkscript-backend-1  |     from flask import Flask
veganetworkscript-backend-1  |   File "/usr/local/lib/python3.9/site-packages/flask/__init__.py", line 19, in <module>
veganetworkscript-backend-1  |     from . import json
veganetworkscript-backend-1  |   File "/usr/local/lib/python3.9/site-packages/flask/json/__init__.py", line 15, in <module>
veganetworkscript-backend-1  |     from itsdangerous import json as _json
veganetworkscript-backend-1  | ImportError: cannot import name 'json' from 'itsdangerous' (/usr/local/lib/python3.9/site-packages/itsdangerous/__init__.py)
veganetworkscript-backend-1 exited with code 1

Here are my requirements.txt, vegamain.py, Dockerfile, and docker-compose.yml files:

requirements.txt:

Flask==2.0.0
Flask-SQLAlchemy==2.4.4
SQLAlchemy==1.3.20
Flask-Migrate==2.5.3
Flask-Script==2.0.6
Flask-Cors==3.0.9
requests==2.25.0
mysqlclient==2.0.1
pika==1.1.0
wolframalpha==4.3.0

vegamain.py:

# Veganetwork (C) TetraSystemSolutions 2022
# all rights are reserved.  
# 
# Author: Trevor R. Blanchard Feb-19-2022-Jul-30-2022
#

# get our imports in order first
from flask import Flask # <-- error occurs here!!!

# start the application through flask.
app = Flask(__name__)

# if set to true will return only a "Hello World" string.
Debug = True

# start a route to the index part of the app in flask.
@app.route('/')
def index():
    if (Debug == True):
        return 'Hello World!'
    else:
        pass

# start the flask app here --->
if __name__ == '__main__':
    app.run(debug=True, host='0.0.0.0')

Dockerfile:

FROM python:3.9
ENV PYTHONUNBUFFERED 1
WORKDIR /app
COPY requirements.txt /app/requirements.txt
RUN pip install -r requirements.txt
COPY . /app

docker-compose.yml:

version: '3.8'
services:
  backend:
    build:
      context: .
      dockerfile: Dockerfile
    command: 'python vegamain.py'
    ports:
      - 8004:5000
    volumes:
      - .:/app
    depends_on:
      - db

#  queue:
#    build:
#      context: .
#      dockerfile: Dockerfile
#    command: 'python -u consumer.py'
#    depends_on:
#      - db

  db:
    image: mysql:5.7.22
    restart: always
    environment:
      MYSQL_DATABASE: admin
      MYSQL_USER: root
      MYSQL_PASSWORD: root
      MYSQL_ROOT_PASSWORD: root
    volumes:
      - .dbdata:/var/lib/mysql
    ports:
      - 33069:3306

How exactly can I fix this code? thank you!

ANSWER

Answered 2022-Feb-20 at 12:31

I was facing the same issue while running docker containers with flask.

I downgraded Flask to 1.1.4 and markupsafe to 2.0.1 which solved my issue.

Check this for reference.

Source https://stackoverflow.com/questions/71189819

QUESTION

Why is it faster to compare strings that match than strings that do not?

Asked 2022-Mar-30 at 11:58

Here are two measurements:

timeit.timeit('"toto"=="1234"', number=100000000)
1.8320042459999968
timeit.timeit('"toto"=="toto"', number=100000000)
1.4517491540000265

As you can see, comparing two strings that match is faster than comparing two strings with the same size that do not match. This is quite disturbing: During a string comparison, I believed that Python was testing strings character by character, so "toto"=="toto" should be longer to test than "toto"=="1234" as it requires four tests against one for the non-matching comparison. Maybe the comparison is hash-based, but in this case, timings should be the same for both comparisons.

Why?

ANSWER

Answered 2022-Mar-30 at 11:57

Combining my comment and the comment by @khelwood:

TL;DR:
When analysing the bytecode for the two comparisons, it reveals the 'time' and 'time' strings are assigned to the same object. Therefore, an up-front identity check (at C-level) is the reason for the increased comparison speed.

The reason for the same object assignment is that, as an implementation detail, CPython interns strings which contain only 'name characters' (i.e. alpha and underscore characters). This enables the object's identity check.

Bytecode:

import dis

In [24]: dis.dis("'time'=='time'")
  1           0 LOAD_CONST               0 ('time')  # <-- same object (0)
              2 LOAD_CONST               0 ('time')  # <-- same object (0)
              4 COMPARE_OP               2 (==)
              6 RETURN_VALUE

In [25]: dis.dis("'time'=='1234'")
  1           0 LOAD_CONST               0 ('time')  # <-- different object (0)
              2 LOAD_CONST               1 ('1234')  # <-- different object (1)
              4 COMPARE_OP               2 (==)
              6 RETURN_VALUE

Assignment Timing:

The 'speed-up' can also be seen in using assignment for the time tests. The assignment (and compare) of two variables to the same string, is faster than the assignment (and compare) of two variables to different strings. Further supporting the hypothesis the underlying logic is performing an object comparison. This is confirmed in the next section.

In [26]: timeit.timeit("x='time'; y='time'; x==y", number=1000000)
Out[26]: 0.0745926329982467

In [27]: timeit.timeit("x='time'; y='1234'; x==y", number=1000000)
Out[27]: 0.10328884399496019

Python source code:

As helpfully provided by @mkrieger1 and @Masklinn in their comments, the source code for unicodeobject.c performs a pointer comparison first and if True, returns immediately.

int
_PyUnicode_Equal(PyObject *str1, PyObject *str2)
{
    assert(PyUnicode_CheckExact(str1));
    assert(PyUnicode_CheckExact(str2));
    if (str1 == str2) {                  // <-- Here
        return 1;
    }
    if (PyUnicode_READY(str1) || PyUnicode_READY(str2)) {
        return -1;
    }
    return unicode_compare_eq(str1, str2);
}

Appendix:

Reference answer nicely illustrating how to read the disassembled bytecode output. Courtesy of @Delgan
Reference answer which nicely describes CPython's string interning. Coutresy of @ShadowRanger

Source https://stackoverflow.com/questions/71644405

QUESTION

Why is `np.sum(range(N))` very slow?

Asked 2022-Mar-29 at 14:31

I saw a video about speed of loops in python, where it was explained that doing sum(range(N)) is much faster than manually looping through range and adding the variables together, since the former runs in C due to built-in functions being used, while in the latter the summation is done in (slow) python. I was curious what happens when adding numpy to the mix. As I expected np.sum(np.arange(N)) is the fastest, but sum(np.arange(N)) and np.sum(range(N)) are even slower than doing the naive for loop.

Why is this?

Here's the script I used to test, some comments about the supposed cause of slowing done where I know (taken mostly from the video) and the results I got on my machine (python 3.10.0, numpy 1.21.2):

updated script:

import numpy as np
from timeit import timeit

N = 10_000_000
repetition = 10

def sum0(N = N):
    s = 0
    i = 0
    while i < N: # condition is checked in python
        s += i
        i += 1 # both additions are done in python
    return s

def sum1(N = N):
    s = 0
    for i in range(N): # increment in C
        s += i # addition in python
    return s

def sum2(N = N):
    return sum(range(N)) # everything in C

def sum3(N = N):
    return sum(list(range(N)))

def sum4(N = N):
    return np.sum(range(N)) # very slow np.array conversion

def sum5(N = N):
    # much faster np.array conversion
    return np.sum(np.fromiter(range(N),dtype = int))

def sum5v2_(N = N):
    # much faster np.array conversion
    return np.sum(np.fromiter(range(N),dtype = np.int_))

def sum6(N = N):
    # possibly slow conversion to Py_long from np.int
    return sum(np.arange(N))

def sum7(N = N):
    # list returns a list of np.int-s
    return sum(list(np.arange(N)))

def sum7v2(N = N):
    # tolist conversion to python int seems faster than the implicit conversion
    # in sum(list()) (tolist returns a list of python int-s)
    return sum(np.arange(N).tolist())

def sum8(N = N):
    return np.sum(np.arange(N)) # everything in numpy (fortran libblas?)

def sum9(N = N):
    return np.arange(N).sum() # remove dispatch overhead

def array_basic(N = N):
    return np.array(range(N))

def array_dtype(N = N):
    return np.array(range(N),dtype = np.int_)

def array_iter(N = N):
    # np.sum's source code mentions to use fromiter to convert from generators
    return np.fromiter(range(N),dtype = np.int_)

print(f"while loop:         {timeit(sum0, number = repetition)}")
print(f"for loop:           {timeit(sum1, number = repetition)}")
print(f"sum_range:          {timeit(sum2, number = repetition)}")
print(f"sum_rangelist:      {timeit(sum3, number = repetition)}")
print(f"npsum_range:        {timeit(sum4, number = repetition)}")
print(f"npsum_iterrange:    {timeit(sum5, number = repetition)}")
print(f"npsum_iterrangev2:  {timeit(sum5, number = repetition)}")
print(f"sum_arange:         {timeit(sum6, number = repetition)}")
print(f"sum_list_arange:    {timeit(sum7, number = repetition)}")
print(f"sum_arange_tolist:  {timeit(sum7v2, number = repetition)}")
print(f"npsum_arange:       {timeit(sum8, number = repetition)}")
print(f"nparangenpsum:      {timeit(sum9, number = repetition)}")
print(f"array_basic:        {timeit(array_basic, number = repetition)}")
print(f"array_dtype:        {timeit(array_dtype, number = repetition)}")
print(f"array_iter:         {timeit(array_iter,  number = repetition)}")

print(f"npsumarangeREP:     {timeit(lambda : sum8(N/1000), number = 100000*repetition)}")
print(f"npsumarangeREP:     {timeit(lambda : sum9(N/1000), number = 100000*repetition)}")

# Example output:
#
# while loop:         11.493371912998555
# for loop:           7.385945574002108
# sum_range:          2.4605720699983067
# sum_rangelist:      4.509678105998319
# npsum_range:        11.85120212900074
# npsum_iterrange:    4.464334709002287
# npsum_iterrangev2:  4.498494338993623
# sum_arange:         9.537815956995473
# sum_list_arange:    13.290120724996086
# sum_arange_tolist:  5.231948580003518
# npsum_arange:       0.241889145996538
# nparangenpsum:      0.21876695199898677
# array_basic:        11.736577274998126
# array_dtype:        8.71628468400013
# array_iter:         4.303306431000237
# npsumarangeREP:     21.240833958996518
# npsumarangeREP:     16.690092379001726

ANSWER

Answered 2021-Oct-16 at 17:42

From the cpython source code for sum sum initially seems to attempt a fast path that assumes all inputs are the same type. If that fails it will just iterate:

/* Fast addition by keeping temporary sums in C instead of new Python objects.
   Assumes all inputs are the same type.  If the assumption fails, default
   to the more general routine.
*/

I'm not entirely certain what is happening under the hood, but it is likely the repeated creation/conversion of C types to Python objects that is causing these slow-downs. It's worth noting that both sum and range are implemented in C.

This next bit is not really an answer to the question, but I wondered if we could speed up sum for python ranges as range is quite a smart object.

To do this I've used functools.singledispatch to override the built-in sum function specifically for the range type; then implemented a small function to calculate the sum of an arithmetic progression.

from functools import singledispatch

def sum_range(range_, /, start=0):
    """Overloaded `sum` for range, compute arithmetic sum"""
    n = len(range_)
    if not n:
        return start
    return int(start + (n * (range_[0] + range_[-1]) / 2))

sum = singledispatch(sum)
sum.register(range, sum_range)

def test():
    """
    >>> sum(range(0, 100))
    4950
    >>> sum(range(0, 10, 2))
    20
    >>> sum(range(0, 9, 2))
    20
    >>> sum(range(0, -10, -1))
    -45
    >>> sum(range(-10, 10))
    -10
    >>> sum(range(-1, -100, -2))
    -2500
    >>> sum(range(0, 10, 100))
    0
    >>> sum(range(0, 0))
    0
    >>> sum(range(0, 100), 50)
    5000
    >>> sum(range(0, 0), 10)
    10
    """

if __name__ == "__main__":
    import doctest
    doctest.testmod()

I'm not sure if this is complete, but it's definitely faster than looping.

Source https://stackoverflow.com/questions/69584027

QUESTION

Error while downloading the requirements using pip install (setup command: use_2to3 is invalid.)

Asked 2022-Mar-05 at 07:13

version pip 21.2.4 python 3.6

The command:

pip install -r  requirments.txt

The content of my requirements.txt:

mongoengine==0.19.1
numpy==1.16.2
pylint
pandas==1.1.5
fawkes

The command is failing with this error

ERROR: Command errored out with exit status 1:
     command: /Users/*/Desktop/ml/*/venv/bin/python -c 'import io, os, sys, setuptools, tokenize; sys.argv[0] = '"'"'/private/var/folders/kn/0y92g7x55qs7c42tln4gwhtm0000gp/T/pip-install-soh30mel/mongoengine_89e68f8427244f1bb3215b22f77a619c/setup.py'"'"'; __file__='"'"'/private/var/folders/kn/0y92g7x55qs7c42tln4gwhtm0000gp/T/pip-install-soh30mel/mongoengine_89e68f8427244f1bb3215b22f77a619c/setup.py'"'"';f = getattr(tokenize, '"'"'open'"'"', open)(__file__) if os.path.exists(__file__) else io.StringIO('"'"'from setuptools import setup; setup()'"'"');code = f.read().replace('"'"'\r\n'"'"', '"'"'\n'"'"');f.close();exec(compile(code, __file__, '"'"'exec'"'"'))' egg_info --egg-base /private/var/folders/kn/0y92g7x55qs7c42tln4gwhtm0000gp/T/pip-pip-egg-info-97994d6e
         cwd: /private/var/folders/kn/0y92g7x55qs7c42tln4gwhtm0000gp/T/pip-install-soh30mel/mongoengine_89e68f8427244f1bb3215b22f77a619c/
    Complete output (1 lines):
    error in mongoengine setup command: use_2to3 is invalid.
    ----------------------------------------
WARNING: Discarding https://*/pypi/packages/mongoengine-0.19.1.tar.gz#md5=68e613009f6466239158821a102ac084 (from https://*/pypi/simple/mongoengine/). Command errored out with exit status 1: python setup.py egg_info Check the logs for full command output.
ERROR: Could not find a version that satisfies the requirement mongoengine==0.19.1 (from versions: 0.15.0, 0.19.1)
ERROR: No matching distribution found for mongoengine==0.19.1

ANSWER

Answered 2021-Nov-19 at 13:30

It looks like setuptools>=58 breaks support for use_2to3:

setuptools changelog for v58

So you should update setuptools to setuptools<58 or avoid using packages with use_2to3 in the setup parameters.

I was having the same problem, pip==19.3.1

Source https://stackoverflow.com/questions/69100275

QUESTION

Repeatedly removing the maximum average subarray

Asked 2022-Feb-28 at 18:19

I have an array of positive integers. For example:

[1, 7, 8, 4, 2, 1, 4]

A "reduction operation" finds the array prefix with the highest average, and deletes it. Here, an array prefix means a contiguous subarray whose left end is the start of the array, such as [1] or [1, 7] or [1, 7, 8] above. Ties are broken by taking the longer prefix.

Original array:  [  1,   7,   8,   4,   2,   1,   4]

Prefix averages: [1.0, 4.0, 5.3, 5.0, 4.4, 3.8, 3.9]

-> Delete [1, 7, 8], with maximum average 5.3
-> New array -> [4, 2, 1, 4]

I will repeat the reduction operation until the array is empty:

[1, 7, 8, 4, 2, 1, 4]
^       ^
[4, 2, 1, 4]
^ ^
[2, 1, 4]
^       ^
[]

Now, actually performing these array modifications isn't necessary; I'm only looking for the list of lengths of prefixes that would be deleted by this process, for example, [3, 1, 3] above.

What is an efficient algorithm for computing these prefix lengths?

The naive approach is to recompute all sums and averages from scratch in every iteration for an O(n^2) algorithm-- I've attached Python code for this below. I'm looking for any improvement on this approach-- most preferably, any solution below O(n^2), but an algorithm with the same complexity but better constant factors would also be helpful.

Here are a few of the things I've tried (without success):

Dynamically maintaining prefix sums, for example with a Binary Indexed Tree. While I can easily update prefix sums or find a maximum prefix sum in O(log n) time, I haven't found any data structure which can update the average, as the denominator in the average is changing.
Reusing the previous 'rankings' of prefix averages-- these rankings can change, e.g. in some array, the prefix ending at index 5 may have a larger average than the prefix ending at index 6, but after removing the first 3 elements, now the prefix ending at index 2 may have a smaller average than the one ending at 3.
Looking for patterns in where prefixes end; for example, the rightmost element of any max average prefix is always a local maximum in the array, but it's not clear how much this helps.

This is a working Python implementation of the naive, quadratic method:

from fractions import Fraction
def find_array_reductions(nums: List[int]) -> List[int]:
    """Return list of lengths of max average prefix reductions."""

    def max_prefix_avg(arr: List[int]) -> Tuple[float, int]:
        """Return value and length of max average prefix in arr."""
        if len(arr) == 0:
            return (-math.inf, 0)

        best_length = 1
        best_average = Fraction(0, 1)
        running_sum = 0

        for i, x in enumerate(arr, 1):
            running_sum += x
            new_average = Fraction(running_sum, i)
            if new_average >= best_average:
                best_average = new_average
                best_length = i

        return (float(best_average), best_length)

    removed_lengths = []
    total_removed = 0

    while total_removed < len(nums):
        _, new_removal = max_prefix_avg(nums[total_removed:])
        removed_lengths.append(new_removal)
        total_removed += new_removal

    return removed_lengths

Edit: The originally published code had a rare error with large inputs from using Python's math.isclose() with default parameters for floating point comparison, rather than proper fraction comparison. This has been fixed in the current code. An example of the error can be found at this Try it online link, along with a foreword explaining exactly what causes this bug, if you're curious.

ANSWER

Answered 2022-Feb-27 at 22:44

This problem has a fun O(n) solution.

If you draw a graph of cumulative sum vs index, then:

The average value in the subarray between any two indexes is the slope of the line between those points on the graph.

The first highest-average-prefix will end at the point that makes the highest angle from 0. The next highest-average-prefix must then have a smaller average, and it will end at the point that makes the highest angle from the first ending. Continuing to the end of the array, we find that...

These segments of highest average are exactly the segments in the upper convex hull of the cumulative sum graph.

Find these segments using the monotone chain algorithm. Since the points are already sorted, it takes O(n) time.

# Lengths of the segments in the upper convex hull
# of the cumulative sum graph
def upperSumHullLengths(arr):
    if len(arr) < 2:
        if len(arr) < 1:
            return []
        else:
            return [1]
    
    hull = [(0, 0),(1, arr[0])]
    for x in range(2, len(arr)+1):
        # this has x coordinate x-1
        prevPoint = hull[len(hull) - 1]
        # next point in cumulative sum
        point = (x, prevPoint[1] + arr[x-1])
        # remove points not on the convex hull
        while len(hull) >= 2:
            p0 = hull[len(hull)-2]
            dx0 = prevPoint[0] - p0[0]
            dy0 = prevPoint[1] - p0[1]
            dx1 = x - prevPoint[0]
            dy1 = point[1] - prevPoint[1]
            if dy1*dx0 < dy0*dx1:
                break
            hull.pop()
            prevPoint = p0
        hull.append(point)
    
    return [hull[i+1][0] - hull[i][0] for i in range(0, len(hull)-1)]


print(upperSumHullLengths([  1,   7,   8,   4,   2,   1,   4]))

prints:

[3, 1, 3]

Source https://stackoverflow.com/questions/71287550

QUESTION

How do I calculate square root in Python?

Asked 2022-Feb-17 at 03:40

I need to calculate the square root of some numbers, for example √9 = 3 and √2 = 1.4142. How can I do it in Python?

The inputs will probably be all positive integers, and relatively small (say less than a billion), but just in case they're not, is there anything that might break?

Related

_{Note: This is an attempt at a canonical question after a discussion on Meta about an existing question with the same title.}

ANSWER

Answered 2022-Feb-04 at 19:44

Option 1: math.sqrt()

The math module from the standard library has a sqrt function to calculate the square root of a number. It takes any type that can be converted to float (which includes int) as an argument and returns a float.

>>> import math
>>> math.sqrt(9)
3.0

Option 2: Fractional exponent

The power operator (**) or the built-in pow() function can also be used to calculate a square root. Mathematically speaking, the square root of a equals a to the power of 1/2.

The power operator requires numeric types and matches the conversion rules for binary arithmetic operators, so in this case it will return either a float or a complex number.

>>> 9 ** (1/2)
3.0
>>> 9 ** .5  # Same thing
3.0
>>> 2 ** .5
1.4142135623730951

(Note: in Python 2, 1/2 is truncated to 0, so you have to force floating point arithmetic with 1.0/2 or similar. See Why does Python give the "wrong" answer for square root?)

This method can be generalized to nth root, though fractions that can't be exactly represented as a float (like 1/3 or any denominator that's not a power of 2) may cause some inaccuracy:

>>> 8 ** (1/3)
2.0
>>> 125 ** (1/3)
4.999999999999999

Edge cases Negative and complex

Exponentiation works with negative numbers and complex numbers, though the results have some slight inaccuracy:

>>> (-25) ** .5  # Should be 5j
(3.061616997868383e-16+5j)
>>> 8j ** .5  # Should be 2+2j
(2.0000000000000004+2j)

Note the parentheses on -25! Otherwise it's parsed as -(25**.5) because exponentiation is more tightly binding than unary negation.

Meanwhile, math is only built for floats, so for x<0, math.sqrt() will raise ValueError: math domain error and for complex x, it'll raise TypeError: can't convert complex to float. Instead, you can use cmath.sqrt(), which is more more accurate than exponentiation (and will likely be faster too):

>>> import cmath
>>> cmath.sqrt(-25)
5j
>>> cmath.sqrt(8j)
(2+2j)

Precision

Both options involve an implicit conversion to float, so floating point precision is a factor. For example:

>>> n = 10**30
>>> square = n**2
>>> x = square**.5
>>> x == n
False
>>> x - n  # how far off are they?
0.0
>>> int(x) - n  # how far off is the float from the int?
19884624838656

Very large numbers might not even fit in a float and you'll get OverflowError: int too large to convert to float. See Python sqrt limit for very large numbers?

Other types

Let's look at Decimal for example:

Exponentiation fails unless the exponent is also Decimal:

>>> decimal.Decimal('9') ** .5
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ** or pow(): 'decimal.Decimal' and 'float'
>>> decimal.Decimal('9') ** decimal.Decimal('.5')
Decimal('3.000000000000000000000000000')

Meanwhile, math and cmath will silently convert their arguments to float and complex respectively, which could mean loss of precision.

decimal also has its own .sqrt(). See also calculating n-th roots using Python 3's decimal module

Source https://stackoverflow.com/questions/70793490

QUESTION

pip-compile raising AssertionError on its logging handler

Asked 2022-Feb-13 at 12:37

I have a dockerfile that currently only installs pip-tools

FROM python:3.9

RUN pip install --upgrade pip && \
    pip install pip-tools

COPY ./ /root/project

WORKDIR /root/project

ENTRYPOINT ["tail", "-f", "/dev/null"]

I build and open a shell in the container using the following commands:

docker build -t brunoapi_image .
docker run --rm -ti --name brunoapi_container --entrypoint bash brunoapi_image

Then, when I try to run pip-compile inside the container I get this very weird error (full traceback):

root@727f1f38f095:~/project# pip-compile
Traceback (most recent call last):
  File "/usr/local/bin/pip-compile", line 8, in <module>
    sys.exit(cli())
  File "/usr/local/lib/python3.9/site-packages/click/core.py", line 1128, in __call__
    return self.main(*args, **kwargs)
  File "/usr/local/lib/python3.9/site-packages/click/core.py", line 1053, in main
    rv = self.invoke(ctx)
  File "/usr/local/lib/python3.9/site-packages/click/core.py", line 1395, in invoke
    return ctx.invoke(self.callback, **ctx.params)
  File "/usr/local/lib/python3.9/site-packages/click/core.py", line 754, in invoke
    return __callback(*args, **kwargs)
  File "/usr/local/lib/python3.9/site-packages/click/decorators.py", line 26, in new_func
    return f(get_current_context(), *args, **kwargs)
  File "/usr/local/lib/python3.9/site-packages/piptools/scripts/compile.py", line 342, in cli
    repository = PyPIRepository(pip_args, cache_dir=cache_dir)
  File "/usr/local/lib/python3.9/site-packages/piptools/repositories/pypi.py", line 106, in __init__
    self._setup_logging()
  File "/usr/local/lib/python3.9/site-packages/piptools/repositories/pypi.py", line 455, in _setup_logging
    assert isinstance(handler, logging.StreamHandler)
AssertionError

I have no clue what's going on and I've never seen this error before. Can anyone shed some light into this?

Running on macOS Monterey

ANSWER

Answered 2022-Feb-05 at 16:30

It is a bug, you can downgrade using:

pip install "pip<22"

https://github.com/jazzband/pip-tools/issues/1558

Source https://stackoverflow.com/questions/70946286

QUESTION

How did print(*a, a.pop(0)) change?

Asked 2022-Feb-04 at 21:21

This code:

a = [1, 2, 3]
print(*a, a.pop(0))

Python 3.8 prints 2 3 1 (does the pop before unpacking).
Python 3.9 prints 1 2 3 1 (does the pop after unpacking).

What caused the change? I didn't find it in the changelog.

Edit: Not just in function calls but also for example in a list display:

a = [1, 2, 3]
b = [*a, a.pop(0)]
print(b)

Prints [2, 3, 1] vs [1, 2, 3, 1]. And Expression lists says "The expressions are evaluated from left to right" (that's the link to Python 3.8 documentation), so I'd expect the unpacking expression to happen first.

ANSWER

Answered 2022-Feb-04 at 21:21

I suspect this may have been an accident, though I prefer the new behavior.

The new behavior is a consequence of a change to how the bytecode for * arguments works. The change is in the changelog under Python 3.9.0 alpha 3:

bpo-39320: Replace four complex bytecodes for building sequences with three simpler ones.

The following four bytecodes have been removed:

BUILD_LIST_UNPACK

BUILD_TUPLE_UNPACK

BUILD_SET_UNPACK

BUILD_TUPLE_UNPACK_WITH_CALL

The following three bytecodes have been added:

LIST_TO_TUPLE

LIST_EXTEND

SET_UPDATE

On Python 3.8, the bytecode for f(*a, a.pop()) looks like this:

  1           0 LOAD_NAME                0 (f)
              2 LOAD_NAME                1 (a)
              4 LOAD_NAME                1 (a)
              6 LOAD_METHOD              2 (pop)
              8 CALL_METHOD              0
             10 BUILD_TUPLE              1
             12 BUILD_TUPLE_UNPACK_WITH_CALL     2
             14 CALL_FUNCTION_EX         0
             16 RETURN_VALUE

while on 3.9, it looks like this:

  1           0 LOAD_NAME                0 (f)
              2 BUILD_LIST               0
              4 LOAD_NAME                1 (a)
              6 LIST_EXTEND              1
              8 LOAD_NAME                1 (a)
             10 LOAD_METHOD              2 (pop)
             12 CALL_METHOD              0
             14 LIST_APPEND              1
             16 LIST_TO_TUPLE
             18 CALL_FUNCTION_EX         0
             20 RETURN_VALUE

In the old bytecode, the code pushes a and (a.pop(),) onto the stack, then unpacks those two iterables into a tuple. In the new bytecode, the code pushes a list onto the stack, then does l.extend(a) and l.append(a.pop()), then calls tuple(l).

This change has the effect of shifting the unpacking of a to before the pop call, but this doesn't seem to have been deliberate. Looking at bpo-39320, the intent was to simplify the bytecode instructions, not to change the behavior, and the bpo thread has no discussion of behavior changes.

Source https://stackoverflow.com/questions/70404485

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Install python

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You can use python like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.