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() could not solve my problem.

This is a simple code in R and its answer:

x <- c(1:4, 0:5, 11)
x
#[1]  1  2  3  4  0  1  2  3  4  5 11
which(x==2)
# [1] 2 7
min(which(x==2))
# [1] 2
which.min(x)
#[1] 5

Which simply returns the index of the item which meets the condition.

If x be the input for Python, how can I get the indeces for the elements which meet criteria x==2 and the one which is the smallest in the array which.min.

x = [1,2,3,4,0,1,2,3,4,11] 
x=np.array(x)
x[x>2].index()
##'numpy.ndarray' object has no attribute 'index'

Asked By: Hadij

||

Source

Answer 1

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

Answered By: Netwave

Answer 2

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

Answered By: erocoar

Answer 3

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]

Answered By: Karl Anka

Answer 4

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]),)

Answered By: 989

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']

Answered By: Ram Prajapati

Answer 6

built-in index function can be used for this purpose:

x = [1,2,3,4,0,1,2,3,4,11] 
print(x.index(min(x)))
#4
print(x.index(max(x)))
#9

However, for indexes based on a condition, np.where or manual loop and enumerate may work:

index_greater_than_two1 = [idx for idx, val in enumerate(x) if val>2]
print(index_greater_than_two1)
# [2, 3, 7, 8, 9]

# OR

index_greater_than_two2 = np.where(np.array(x)>2)
print(index_greater_than_two2)
# (array([2, 3, 7, 8, 9], dtype=int64),)

Answered By: Hadij

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

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