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borel | Enables regular set operations on any ordered class | Date Time Utils library

 by   badosu Ruby Version: Current License: Non-SPDX

 by   badosu Ruby Version: Current License: Non-SPDX

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kandi X-RAY | borel Summary

borel is a Ruby library typically used in Utilities, Date Time Utils applications. borel has no bugs, it has no vulnerabilities and it has low support. However borel has a Non-SPDX License. You can download it from GitHub.
[![Build Status](https://travis-ci.org/badosu/borel.png)](https://travis-ci.org/badosu/borel) [![Gem Version](https://badge.fury.io/rb/borel.png)](http://badge.fury.io/rb/borel) [![Code Climate](https://codeclimate.com/github/badosu/borel.png)](https://codeclimate.com/github/badosu/borel). Borelian sets are formed by enumerable union, intersection or complement, of intervals. Borel enables performing regular operations on intervals of any comparable class. Borel borrows many of the ideas (and code) from the Intervals [gem][1]. However it differs from Intervals in which it’s aim is not on numerical precision and calculation, but on ease of use and solving some general interval related problems.
Support
Support
Quality
Quality
Security
Security
License
License
Reuse
Reuse

kandi-support Support

  • borel has a low active ecosystem.
  • It has 7 star(s) with 2 fork(s). There are 1 watchers for this library.
  • It had no major release in the last 12 months.
  • borel has no issues reported. There are no pull requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of borel is current.
borel Support
Best in #Date Time Utils
Average in #Date Time Utils
borel Support
Best in #Date Time Utils
Average in #Date Time Utils

quality kandi Quality

  • borel has 0 bugs and 0 code smells.
borel Quality
Best in #Date Time Utils
Average in #Date Time Utils
borel Quality
Best in #Date Time Utils
Average in #Date Time Utils

securitySecurity

  • borel has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
  • borel code analysis shows 0 unresolved vulnerabilities.
  • There are 0 security hotspots that need review.
borel Security
Best in #Date Time Utils
Average in #Date Time Utils
borel Security
Best in #Date Time Utils
Average in #Date Time Utils

license License

  • borel has a Non-SPDX License.
  • Non-SPDX licenses can be open source with a non SPDX compliant license, or non open source licenses, and you need to review them closely before use.
borel License
Best in #Date Time Utils
Average in #Date Time Utils
borel License
Best in #Date Time Utils
Average in #Date Time Utils

buildReuse

  • borel releases are not available. You will need to build from source code and install.
  • Installation instructions, examples and code snippets are available.
  • It has 668 lines of code, 43 functions and 26 files.
  • It has low code complexity. Code complexity directly impacts maintainability of the code.
borel Reuse
Best in #Date Time Utils
Average in #Date Time Utils
borel Reuse
Best in #Date Time Utils
Average in #Date Time Utils
Top functions reviewed by kandi - BETA

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

  • returns a random range

borel Key Features

Enables regular set operations on any ordered class

Install

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$ gem install borel

Initializing

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  Interval[]
  Interval[1]
  Interval[0,1]
  Interval[[0,1],[2,3],[5]]

Properties

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  Interval[1,2].inf             # -> 1
  Interval[1,2].sup             # -> 2
  Interval[1].degenerate?       # -> true
  Interval[[0,1],[2,3]].simple? # -> false
  Interval[].empty?             # -> true
  Interval[1,5].include?(3.4)   # -> true

Operations

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    ~Interval[0,5]             # -> Interval[[-Infinity, 0], [5, Infinity]]

Generic Intervals

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Interval['a','c'] ^ Interval['b','d'] # -> Interval['b','c']
Interval['a','c'] | Interval['b','d'] # -> Interval['a','d']

Math Extensions

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require 'borel/math_extensions'

Interval[1,5].rand  # -> Random.new.rand 1..5
Interval[1,5].width # -> 5-1, only for simple intervals

Convert only the numeric elements of pd.series to float

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def convert_to_number(s): 
    n = s.split('  ').strip()  # work with the right-hand portion only 
    if 'milhões' in n: 
        return 1e6 * float(n.split()[0]) 
    if 'mil' in n: 
        return 1e3 * float(n.split()[0]) 
    if 'zero' in n: 
        return 0 
    return np.nan
new_series = series.apply(convert_to_number)
-----------------------
def convert_to_number(s): 
    n = s.split('  ').strip()  # work with the right-hand portion only 
    if 'milhões' in n: 
        return 1e6 * float(n.split()[0]) 
    if 'mil' in n: 
        return 1e3 * float(n.split()[0]) 
    if 'zero' in n: 
        return 0 
    return np.nan
new_series = series.apply(convert_to_number)
-----------------------
import pandas as pd
import numpy as np

data = """\
Douglas Friedrich         custo zero
Ernando                   custo zero
Lucas Fonseca             custo zero
Éverson                           NA
Juninho Capixaba        desconhecido
Zeca                    desconhecido
Nino Paraíba              custo zero
Douglas Borel                     NA
Elton                     custo zero
Ronaldo                   custo zero
Jádson                  desconhecido
Rodriguinho               custo zero
Marco Antônio              120 mil €
Clayson                    880 mil €
Rossi                     custo zero
Gilberto                  custo zero
Saldanha                          NA
Anderson                          NA
Juninho               1.30 milhões €
Wanderson                  400 mil €
Ignácio                 desconhecido
Giovanni                  custo zero
João Pedro              desconhecido
Matheus Silva             custo zero
Gregore                    235 mil €
Flávio                  desconhecido
Yuri                      custo zero
Saldanha                          NA
Danielzinho               custo zero
Alesson                 desconhecido
Élber                      130 mil €
Gustavo                           NA
Fernandão             1.06 milhões €"""

data = pd.Series(
    data.split('\n')
).str.replace(
    '  +', '|'
).str.rsplit(
    '|', expand=True
).rename(
    columns=dict(zip(range(2), ['who', 'value']))
)

f = data.value.str.contains('milhões €')
data.loc[f, 'num'] = data.loc[f, 'value'].str.split(' ', 1).str[0].apply(float) * 1000000
f = data.value.str.contains('mil €')
data.loc[f, 'num'] = data.loc[f, 'value'].str.split(' ', 1).str[0].apply(float) * 1000

Community Discussions

Trending Discussions on borel
  • Convert only the numeric elements of pd.series to float
Trending Discussions on borel

QUESTION

Convert only the numeric elements of pd.series to float

Asked 2020-Jul-02 at 14:58
nome
Douglas Friedrich         custo zero
Ernando                   custo zero
Lucas Fonseca             custo zero
Éverson                           NA
Juninho Capixaba        desconhecido
Zeca                    desconhecido
Nino Paraíba              custo zero
Douglas Borel                     NA
Elton                     custo zero
Ronaldo                   custo zero
Jádson                  desconhecido
Rodriguinho               custo zero
Marco Antônio              120 mil €
Clayson                    880 mil €
Rossi                     custo zero
Gilberto                  custo zero
Saldanha                          NA
Anderson                          NA
Juninho               1.30 milhões €
Wanderson                  400 mil €
Ignácio                 desconhecido
Giovanni                  custo zero
João Pedro              desconhecido
Matheus Silva             custo zero
Gregore                    235 mil €
Flávio                  desconhecido
Yuri                      custo zero
Saldanha                          NA
Danielzinho               custo zero
Alesson                 desconhecido
Élber                      130 mil €
Gustavo                           NA
Fernandão             1.06 milhões €
Name: preco_pago, dtype: object

Is there anyway that I can convert only the numeric element of this series to float(ex: 1.30 milhões €) and leave the rest as it is ? I also would like to convert thousands to 1000 and millions to millions 1000000. This is what I actually want : 1.30 milhões € to 1300000

ANSWER

Answered 2020-Jul-02 at 14:52

If I understood correctly, you have a pandas Series with strings (i.e. your first element is the string 'Douglas Friedrich custo zero'.

You'll have to create your own function to parse each string, for example:

def convert_to_number(s): 
    n = s.split('  ').strip()  # work with the right-hand portion only 
    if 'milhões' in n: 
        return 1e6 * float(n.split()[0]) 
    if 'mil' in n: 
        return 1e3 * float(n.split()[0]) 
    if 'zero' in n: 
        return 0 
    return np.nan

Then simply apply this function to your pandas series:

new_series = series.apply(convert_to_number)

Of course, you may need to modify the function depending on your needs, or adapt it to any edge cases that may come along.

Source https://stackoverflow.com/questions/62698871

Community Discussions, Code Snippets contain sources that include Stack Exchange Network

Vulnerabilities

No vulnerabilities reported

Install borel

You may install it traditionally, tipically for interactive sessions:. Or just put this somewhere on your application’s Gemfile.

Support

For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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