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Random numbers Continuous in python


Random numbers Continuous in python

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Random numbers Continuous in python
Tag : python , By : qba73
Date : November 29 2020, 12:01 PM

I hope this helps . Use random.choices as:
random.choices(range(10), k=10)
[9, 7, 3, 3, 2, 1, 0, 4, 2, 4]

int(''.join(map(str, random.choices(range(10), k=10))))
3853229930

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Given a set of random numbers drawn from a continuous univariate distribution, find the distribution


Tag : r , By : Nickolas
Date : March 29 2020, 07:55 AM
it helps some times My first approach would be to generate qq plots of the given data against the possible distributions.
x <- c(15.771062,14.741310,9.081269,11.276436,11.534672,17.980860,13.550017,13.853336,11.262280,11.049087,14.752701,4.481159,11.680758,11.451909,10.001488,11.106817,7.999088,10.591574,8.141551,12.401899,11.215275,13.358770,8.388508,11.875838,3.137448,8.675275,17.381322,12.362328,10.987731,7.600881,14.360674,5.443649,16.024247,11.247233,9.549301,9.709091,13.642511,10.892652,11.760685,11.717966,11.373979,10.543105,10.230631,9.918293,10.565087,8.891209,10.021141,9.152660,10.384917,8.739189,5.554605,8.575793,12.016232,10.862214,4.938752,14.046626,5.279255,11.907347,8.621476,7.933702,10.799049,8.567466,9.914821,7.483575,11.098477,8.033768,10.954300,8.031797,14.288100,9.813787,5.883826,7.829455,9.462013,9.176897,10.153627,4.922607,6.818439,9.480758,8.166601,12.017158,13.279630,14.464876,13.319124,12.331335,3.194438,9.866487,11.337083,8.958164,8.241395,4.289313,5.508243,4.737891,7.577698,9.626720,16.558392,10.309173,11.740863,8.761573,7.099866,10.032640)
> qqnorm(x)
> library(MASS)
> fitdistr(x, 't')$loglik
[1] -252.2659
Warning message:
In log(s) : NaNs produced
> fitdistr(x, 'normal')$loglik
[1] -252.2968
> fitdistr(x, 'logistic')$loglik
[1] -252.2996
> fitdistr(x, 'weibull')$loglik
[1] -252.3507
> fitdistr(x, 'gamma')$loglik
[1] -255.9099
> fitdistr(x, 'lognormal')$loglik
[1] -260.6328
> fitdistr(x, 'exponential')$loglik
[1] -331.8191
Warning messages:
1: In dgamma(x, shape, scale, log) : NaNs produced
2: In dgamma(x, shape, scale, log) : NaNs produced

write a c function that generates one random number, or a pair of random numbers, or a triplet of random numbers given t


Tag : c , By : Tamizhvendan
Date : March 29 2020, 07:55 AM
With these it helps If the range is small enough, you shouldn't have problems in using the usual modulo method
int GetRandomInt(int Min, int Max)
{
    return (rand()%(Max-Min+1))+Min;
}
int GetRandomInt(int Min, int Max)
{
    return (int)(((double)rand())/MAX_RAND*(Max-Min))+Min;
}

Extract a number of continuous digits from a random string in python


Tag : python , By : ponchopilate
Date : March 29 2020, 07:55 AM
I wish did fix the issue. I am trying to parse this list of strings that contains ID values as a series of 7 digits but I am not sure how to approach this. , You can use regular expression, like this
import re
pattern = re.compile(r"\bid\s*?(\d+)")
print ["; ".join(pattern.findall(item)) for item in lst1]
# ['3925810; 3925721', '3976019; 3975995']
pattern = re.compile(r"\bid\s*?(\d{7})\D*?")

How to make continuous random NaNs in python array


Tag : python , By : Andrew L.
Date : March 29 2020, 07:55 AM
hop of those help? I can suggest this approach, which is far from being optimal, but it works:
## data
import numpy as np
np.random.seed(42)

a = np.array([[1., 3., 4., 5., 1., 2., 5., 7.],
            [1., 3., 6., 4., 4., 0., 4., 0.],
            [5., 0., 5., 1., 3., 0., 5., 1.],
            [2., 1., 5., 4., 6., 1., 3., 4.],
            [2., 1., 0., 0., 6., 1., 2., 1.],
            [3., 5., 7., 2., 9., 6., 3., 5.]])

## NaN statistics definition
nan_ratio = 0.2
nan_seq_len = 4
number_of_nan_seq = int(nan_ratio * a.size / nan_seq_len)
nan_position = np.random.randint(0, a.size-nan_seq_len, size=number_of_nan_seq)

## NaN insertion
nan_indexes = np.ones([nan_position.size, nan_seq_len]) * nan_position[:, np.newaxis] + np.arange(0, nan_seq_len)
a_shape = a.shape
a_flat = a.flatten()
a_flat[nan_indexes.astype('int')] = np.nan
a = a_flat.reshape(a_shape)
array([[ 1.,  3.,  4.,  5.,  1.,  2.,  5.,  7.],
       [ 1.,  3.,  6.,  4.,  4.,  0.,  4.,  0.],
       [ 5.,  0.,  5.,  1.,  3.,  0.,  5.,  1.],
       [ 2.,  1.,  5.,  4., nan, nan, nan, nan],
       [ 2.,  1.,  0.,  0.,  6.,  1., nan, nan],
       [nan, nan,  7.,  2.,  9.,  6.,  3.,  5.]])

How do I print the total of random numbers without making the random numbers variables in Python?


Tag : python , By : LinnheCreative
Date : March 29 2020, 07:55 AM
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