CODEWITHSUNDEEP
pythonpandas
Harshavardhan Ramanna
Harshavardhan Ramanna

Reputation: 738

Randomly choose two values without repetition in dataframe

Consider a dataframe df with N columns and M rows:

>>> df = pd.DataFrame(np.random.randint(1, 10, (10, 5)), columns=list('abcde'))
>>> df
   a  b  c  d  e
0  4  4  5  5  7
1  9  3  8  8  1
2  2  8  1  8  5
3  9  5  1  2  7
4  3  5  8  2  3
5  2  8  8  2  8
6  3  1  7  2  6
7  4  1  5  6  3
8  5  4  4  9  5
9  3  7  5  6  6

I want to randomly choose two columns and then randomly choose one particular row (this would give me two values of the same row). I can achieve this using 

>>> df.sample(2, axis=1).sample(1,axis=0)
   e  a
1  3  5

I want to perform this K times like below :

>>> for i in xrange(5):
...     df.sample(2, axis=1).sample(1,axis=0)
...
   e  a
1  3  5
   d  b
2  1  9
   e  b
4  8  9
   c  b
0  6  5
   e  c
1  3  5

I want to ensure that I do not choose the same two values (by choosing the same two columns and same row) in any of the trials. How would I achieve this?

I want to then perform a bitwise XOR operation on the two chosen values in each trial as well. For example, 3 ^ 5, 1 ^ 9 , .. and count all the bit differences in the chosen values.

Upvotes: 2

Views: 674

Answers (1)

ALollz
ALollz

Reputation: 59579

You can create a list of all of the index by 2 column tuples. And then take random selections from that without replacement.

Sample Data

import pandas as pd
import numpy as np
from itertools import combinations, product

np.random.seed(123)
df = pd.DataFrame(np.random.randint(1, 10, (10, 5)), columns=list('abcde'))
#df = df.reset_index() #if index contains duplicates

Code

K = 5
choices = np.array(list(product(df.index, combinations(df.columns, 2))))
idx = choices[np.r_[np.random.choice(len(choices), K, replace=False)]]

#array([[9, ('a', 'e')],
#       [2, ('a', 'e')],
#       [1, ('a', 'c')],
#       [3, ('b', 'e')],
#       [8, ('d', 'e')]], dtype=object)

Then you can decide how exactly you want your output, but something like this is close to what you show:

pd.concat([df.loc[myid[0], list(myid[1])].reset_index().T for myid in idx])
#       0  1
#index  a  e
#9      4  8
#index  a  e
#2      1  1
#index  a  c
#1      7  1
#index  b  e
#3      2  3
#index  d  e
#8      5  7

Upvotes: 4

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