low | low level data type and utils in Golang | REST library
kandi X-RAY | low Summary
kandi X-RAY | low Summary
low level data type and utils in Golang. A stable low level function set is the basis of a robust architecture. It focuses on stability and requires high test coverage.
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Top functions reviewed by kandi - BETA
- Select32 returns a random 32 - bit value pair .
- select32single finds the i - th index of i .
- size returns the total size of v .
- stat returns a string representation of v .
- IndexToPath converts an index to a path .
- ShardByPrefix splits a set of keys by prefixes .
- Select32R64 selects an int32 from a list of words
- FromStr32 returns the length of the string from the string s .
- AllPaths returns all paths of the bitmap from from to to to .
- CmpUpto returns 0 if a is less than b
low Key Features
low Examples and Code Snippets
def create_low_latency_svdf_model(fingerprint_input, model_settings,
is_training, runtime_settings):
"""Builds an SVDF model with low compute requirements.
This is based in the topology presented in the 'Compres
def create_low_latency_conv_model(fingerprint_input, model_settings,
is_training):
"""Builds a convolutional model with low compute requirements.
This is roughly the network labeled as 'cnn-one-fstride4' in the
function loopDetection2() {
var map = new Map();
for(let i = this.head; i; i = i.next) {
if(map.has(i)) {
return i;
} else {
map.set(i, 1);
}
}
return false;
}
Community Discussions
Trending Discussions on low
QUESTION
ANSWER
Answered 2021-Jun-16 at 01:11The problem is that your CSS selectors include parentheses ()
and dollar signs $
. These symbols already have a special meaning. See:
You can escape these characters using a backslash \
.
QUESTION
I am trying to check if the value in 'diff'
column is greater than 0
if it is, then the value in 'worth'
should be False
else it should be True
I am using the below code to compute and check but it always gives me True
. Can anyone point here what is the mistake. I am attaching pic of output as well
ANSWER
Answered 2021-Jun-15 at 20:37Try with subtraction + np.where
instead:
QUESTION
I have bitcoin historical data. I split the "DATE" Colum as "year month days and hour" because I wanted to sort data based on hours["AS it is hourly base data"]. the data goes up to 2021-12 ie Decmber["the dates goes from 1 to 30 every month"]. I want to sort this data further as:- "2019-Jan, 2020-Jan 20201-Jan" then "2019-Feb,2020-Feb, 2021-Feb" and soon on
Year Month DAy Hour open high low close 2019 1 1 0 3700.05 3725.58 3698.83 3715.09 2019 2 1 0 3700.05 3725.58 3698.83 3715.09 2019 3 1 0 3700.05 3725.58 3698.83 3715.09 2019 4 1 0 3700.05 3725.58 3698.83 3715.09 2019 5 1 0 3700.05 3725.58 3698.83 3715.09can this be done by saying I have split the "DATE" column? if YES please any suggestion on how this can be achieved
The original "DATE" colum was as follows :- 2019-01-01T00:00:00Z
...ANSWER
Answered 2021-Jun-15 at 18:58You can first sort with respect to Month
and then Year
:
QUESTION
When a divide-and-conquer recursive function doesn't yield runtimes low enough, which other improvements could be done?
Let's say, for example, this power
function taken from here:
ANSWER
Answered 2021-Jun-15 at 17:36The primary optimization you should use here is common subexpression elimination. Consider your first piece of code:
QUESTION
I have a dataframe output from the python script which gives following output
Datetime High Low Time 546 2021-06-15 14:30:00 15891.049805 15868.049805 14:30:00 547 2021-06-15 14:45:00 15883.000000 15869.900391 14:45:00 548 2021-06-15 15:00:00 15881.500000 15866.500000 15:00:00 549 2021-06-15 15:15:00 15877.750000 15854.549805 15:15:00 550 2021-06-15 15:30:00 15869.250000 15869.250000 15:30:00i Want to remove all rows where time is equal to 15:30:00. tried different things but unable to do. Help please.
...ANSWER
Answered 2021-Jun-15 at 15:55The way I did was the following,
First we get the the time we want to remove from the dataset, that is 15:30:00 in this case.
Since the Datetime column is in the datetime format, we cannot compare the time as strings. So we convert the given time in the datetime.time() format.
rm_time = dt.time(15,30)
With this, we can go about using the DataFrame.drop()
df.drop(df[df.Datetime.dt.time == rm_time].index)
QUESTION
I have a dataset with the name of Danish ministers and their position from 1990 to 2020 (data comes from dataset called WhoGovern; https://politicscentre.nuffield.ox.ac.uk/whogov-dataset/). The dataset consists of the ministers name
, the ministers position
, the prestige
of that position, and the year
in which the minister had that given position.
My problem is that some ministers are counted twice in the same year (i.e., the rows aren't unique in terms of name
and year
). See the example in the picture below, where "Bertel Haarder" was both Minister of Health and Minister of Interior Affairs in 2010 and 2021.
I want to create a dataset, where all the rows are unique combinations of name
and year
. However, I do not want to remove any information from the dataset. Instead, I want to use the information in the prestige
column to combine the duplicated rows into one. The observations with the highest prestige should be the main observations, where the other information should be added in a new column, e.g., position2
and prestige2
. In the example with Bertel Haarder the data should look like this:
(PS: Sorry for bad presenting of the tables, but didn't know how to create a nice looking table...)
Here's the dataset for creating a reproducible example with observations from 2010-2020:
...ANSWER
Answered 2021-Jun-08 at 14:04Reshape the data to wide format twice, once for position
and the other for prestige_1
, and join the two results.
QUESTION
Good day, everyone.
I want to have two separate TensorFlow models (f
and g
) and train both of them on the loss of f(g(x)). However, I want to use them separately, like g(x) or f(e), where e is an embedded vector but received not from g.
For example, the classical way to create the model with embedding looks like this:
...ANSWER
Answered 2021-Jun-15 at 10:53This can be achieved by weight sharing or shared layers. To share layers in different models in keras, you just need to pass the same instance of layer to both of the models.
Example Codes:
QUESTION
Following script is a combination of RSI and Higher High and Lower Low script. The issue is that the HH LL labels are aligned for the price not on the RSI Line. How to align the labels to the RSI line? It is basically showing the Higher Highs and Lower Lows of RSI. The labels need to stick on to the respective RSI line.
...ANSWER
Answered 2021-Jun-15 at 09:25Changed the location.belowbar
and location.abovebar
with location.absolute
and the plotshapes display (ex: if _hl is true, plot at the RSI level, otherwise pass)
QUESTION
I know there are some other questions (with answers) to this topic. But no of these was helpful for me.
I have a postfix server (postfix 3.4.14 on debian 10) with following configuration (only the interesting section):
...ANSWER
Answered 2021-Jun-15 at 08:30Here I'm wondering about the line [in s_client]
New, TLSv1/SSLv3, Cipher is ECDHE-RSA-AES256-GCM-SHA384
You're apparently using OpenSSL 1.0.2, where that's a basically useless relic. Back in the days when OpenSSL supported SSLv2 (mostly until 2010, although almost no one used it much after 2000), the ciphersuite values used for SSLv3 and up (including all TLS, but before 2014 OpenSSL didn't implement higher than TLS1.0) were structured differently than those used for SSLv2, so it was important to qualify the ciphersuite by the 'universe' it existed in. It has almost nothing to do with the protocol version actually used, which appears later in the session-param decode:
QUESTION
I would like to find minimum distance of each voxel to a boundary element in a binary image in which the z voxel size is different from the xy voxel size. This is to say that a single voxel represents a 225x110x110 (zyx) nm volume.
Normally, I would do something with scipy.ndimage.morphology.distance_transform_edt (https://docs.scipy.org/doc/scipy-0.14.0/reference/generated/scipy.ndimage.morphology.distance_transform_edt.html) but this gives the assume that isotropic sizes of the voxel:
...ANSWER
Answered 2021-Jun-15 at 02:32Normally, I would do something with scipy.ndimage.morphology.distance_transform_edt but this gives the assume that isotropic sizes of the voxel:
It does no such thing! You are looking for the sampling=
parameter. From the latest version of the docs:
Spacing of elements along each dimension. If a sequence, must be of length equal to the input rank; if a single number, this is used for all axes. If not specified, a grid spacing of unity is implied.
The wording "sampling" or "spacing" is probably a bit mysterious if you think of pixels as little squares/cubes, and that is probably why you missed it. In most situations, it is better to think of pixels as point samples on a grid, with fixed spacing between samples. I recommend Alvy Ray's a pixel is not a little square for a better understanding of this terminology.
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