If I have:
x = np.array(([1,4], [2,5], [2,6], [3,4], [3,6], [3,7], [4,3], [4,5], [5,2]))
for item in range(3):
choice = random.choice(x)
How can I get the index number of the random choice taken from the array?
I tried:
indexNum = np.where(x == choice)
print(indexNum[0])
But it didn't work.
I want the output, for example, to be something like:
chosenIndices = [1 5 8]
Another possibility is using np.where and np.intersect1d. Here random choice without repetition.
x = np.array(([1,4], [2,5], [2,6], [3,4], [3,6], [3,7], [4,3], [4,5], [5,2]))
res=[]
cont = 0
while cont<3:
choice = random.choice(x)
ind = np.intersect1d(np.where(choice[0]==x[:,0]),np.where(choice[1]==x[:,1]))[0]
if ind not in res:
res.append(ind)
cont+=1
print (res)
# Output [8, 1, 5]
You can achieve this by converting the numpy array to list of tuples and then apply the index function.
This would work:
import random
import numpy as np
chosenIndices = []
x = np.array(([1,4], [2,5], [2,6], [3,4], [3,6], [3,7], [4,3], [4,5], [5,2]))
x = x.T
x = list(zip(x[0],x[1]))
item = 0
while len(chosenIndices)!=3:
choice = random.choice(x)
indexNum = x.index(choice)
if indexNum in chosenIndices: # if index already exist, then it will rerun that particular iteration again.
item-=1
else:
chosenIndices.append(indexNum)
print(chosenIndices) # Thus all different results.
Output:
[1, 3, 2]
