get indexes with condition

Question

How do I find a position(index) of all entries of search_value?

import pandas as pd
import numpy as np
search_value=8
lst=[5, 8, 2, 7, 8, 8, 2, 4]
df=pd.DataFrame(lst)
df["is_search_value"]=np.where(df[0]==search_value, True, False)
print(df.head(20))

Output:

   0  is_search_value
0  5            False
1  8             True
2  2            False
3  7            False
4  8             True
5  8             True
6  2            False
7  4            False

Desirable output:

1 4 5

If search_value is 10 than desirable output:

None

If you don't pass True and False to np.where, it returns indices (`np.where(df[0]==search_value)`). You'll get `(array([1, 4, 5]),)`. — ayhan, Jan 18 '18 at 08:10
@ayhan if I don't pass True and False I'm getting error "ValueError: Length of values does not match length of index" — user2978216, Jan 18 '18 at 08:12
Are you sure? `np.where(df[0]==search_value)` returns `(array([1, 4, 5]),)` in your example. — ayhan, Jan 18 '18 at 08:13

Alexander · Accepted Answer · 2018-01-18T08:10:49.860

3

You can use enumerate in a conditional list comprehension to get the index locations.

my_list = [5, 8, 2, 7, 8, 8, 2, 4]
search_value = 8
>>> [i for i, n in enumerate(my_list) if n == search_value]
[1, 4, 5]

If the search value is not in the list, then an empty list will be returned (not exactly None, but still a falsey).

Using pandas, you can use boolean indexing to get the matches, then extract the index to a list:

df[df[0] == search_value].index.tolist()

Using an empty list will satisfy the condition for None (they both evaluate to False). If you really need None, then use the suggestion of @cᴏʟᴅsᴘᴇᴇᴅ.

edited Jan 18 '18 at 08:10

answered Jan 18 '18 at 08:04

Alexander

105,104
32
201
196

List comprehension? Where are pandas in that? – user2978216 Jan 18 '18 at 08:04
@user2978216 You don't have to shoot a gun when a pistol is more than enough. – Szabolcs Jan 18 '18 at 08:06
3

Heheh, these comments. Also, for OP's requirement of returning None, the condition would be `df.index[df[0] == search_value].tolist() or None`. – cs95 Jan 18 '18 at 08:10
1

@user2978216 the panda's at the bottom next to his username – Daniel F Jan 18 '18 at 12:14

score 1 · Answer 2 · answered Jan 18 '18 at 09:41

If you pass to np.where() only condition - it'll return indices of matched elements:

In [29]: np.where(df[0]==search_value)
Out[29]: (array([1, 4, 5], dtype=int64),)

For custom Pandas indices:

In [39]: df = pd.DataFrame(lst, index=np.arange(8)[::-1])

In [40]: df
Out[40]:
   0
7  5
6  8
5  2
4  7
3  8
2  8
1  2
0  4

In [41]: df.index[np.where(df[0]==search_value)]
Out[41]: Int64Index([6, 3, 2], dtype='int64')

score 0 · Answer 3 · answered Jan 18 '18 at 08:18

0

with pandas

>>> (pd.Series([5, 8, 2, 7, 8, 8, 2, 4]) == 8).nonzero()
(array([1, 4, 5], dtype=int64),)

answered Jan 18 '18 at 08:18

C. Yduqoli

1,706
14
18

@ayhan Yes, my mistake. – Alexander Jan 18 '18 at 08:35

get indexes with condition

3 Answers3