R's which() and which.min() Equivalent in Python

Question

I read the similar topic here. I think the question is different or at least .index() couldnot solve my problem.

This is a simple code in R and its answ

Answer 1

You could also use heapq to find the index of the smallest. Then you can chose to find multiple (for example index of the 2 smallest).

import heapq

x = np.array([1,2,3,4,0,1,2,3,4,11]) 

heapq.nsmallest(2, (range(len(x))), x.take)

Returns [4, 0]

Answer 2

NumPy for R provides you with a bunch of R functionalities in Python.

As to your specific question:

import numpy as np
x = [1,2,3,4,0,1,2,3,4,11] 
arr = np.array(x)
print(arr)
# [ 1  2  3  4  0  1  2  3  4 11]

print(arr.argmin(0)) # R's which.min()
# 4

print((arr==2).nonzero()) # R's which()
# (array([1, 6]),)

Answer 3

Numpy does have built-in functions for it

x = [1,2,3,4,0,1,2,3,4,11] 
x=np.array(x)
np.where(x == 2)
np.min(np.where(x==2))
np.argmin(x)

np.where(x == 2)
Out[9]: (array([1, 6], dtype=int64),)

np.min(np.where(x==2))
Out[10]: 1

np.argmin(x)
Out[11]: 4

Answer 4

A simple loop will do:

res = []
x = [1,2,3,4,0,1,2,3,4,11] 
for i in range(len(x)):
    if check_condition(x[i]):
        res.append(i)

One liner with comprehension:

res = [i for i, v in enumerate(x) if check_condition(v)]

Here you have a live example

Answer 5

The method based on python indexing and numpy, which returns the value of the desired column based on the index of the minimum/maximum value

df.iloc[np.argmin(df['column1'].values)]['column2']