Python/Pandas return column and row index of found string

Question

I've searched previous answers relating to this but those answers seem to utilize numpy because the array contains numbers. I am trying to search for a keyword in a sentence in a dataframe ('Timeframe') where the full sentence is 'Timeframe for wave in ____' and would like to return the column and row index. For example:

    df.iloc[34,0]

returns the string I am looking for but I am avoiding a hard code for dynamic reasons. Is there a way to return the [34,0] when I search the dataframe for the keyword 'Timeframe'

You can access the corresponding row by using df.index.get_loc as explained in the target. — user2285236
– user2285236, Commented Jul 11, 2017 at 18:38
@ayhan - I reopen it, because it seems get_loc is not solution. — jezrael
– jezrael, Commented Jul 11, 2017 at 18:42

jezrael · Accepted Answer · 2017-07-11 19:11:58Z

4

EDIT:

For check index need contains with boolean indexing, but then there are possible 3 values:

df = pd.DataFrame({'A':['Timeframe for wave in ____', 'a', 'c']})
print (df)
                            A
0  Timeframe for wave in ____
1                           a
2                           c



def check(val):
    a = df.index[df['A'].str.contains(val)]
    if a.empty:
        return 'not found'
    elif len(a) > 1:
        return a.tolist()
    else:
        #only one value - return scalar  
        return a.item()

print (check('Timeframe'))
0

print (check('a'))
[0, 1]

print (check('rr'))
not found

Old solution:

It seems you need if need numpy.where for check value Timeframe:

df = pd.DataFrame({'A':list('abcdef'),
                   'B':[4,5,4,5,5,4],
                   'C':[7,8,9,4,2,'Timeframe'],
                   'D':[1,3,5,7,1,0],
                   'E':[5,3,6,9,2,4],
                   'F':list('aaabbb')})

print (df)
   A  B          C  D  E  F
0  a  4          7  1  5  a
1  b  5          8  3  3  a
2  c  4          9  5  6  a
3  d  5          4  7  9  b
4  e  5          2  1  2  b
5  f  4  Timeframe  0  4  b


a = np.where(df.values == 'Timeframe')
print (a)
(array([5], dtype=int64), array([2], dtype=int64))

b = [x[0] for x in a]
print (b)
[5, 2]

edited Jul 11, 2017 at 19:11

answered Jul 11, 2017 at 18:41

jezrael

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3 Comments

codeninja Over a year ago

Thanks jezrael! one thing I didn't realize would be more problematic is that Timeframe is not the whole word. I initially tried doing a df[df == 'Timeframe'] kind of search expecting it to locate the first instance of Timeframe. However since Timeframe is part of a sentence it will not return a result. The full sentence is 'Timeframe for wave in____ ' do you have any tips?

jezrael Over a year ago

Hmm, and output is positions? Or output is all sentence?

codeninja Over a year ago

since my df is only one column, i would just need to return the row where the sentence 'Timeframe for wave in___' appears. Does that help clarify? So in my example df.iloc[34,0] actually corresponds to where the sentence appears

Cedric Zoppolo · Accepted Answer · 2017-07-11 19:18:01Z

3

In case you have multiple columns where to look into you can use following code example:

import numpy as np
import pandas as pd
df = pd.DataFrame([[1,2,3,4],["a","b","Timeframe for wave in____","d"],[5,6,7,8]])
mask = np.column_stack([df[col].str.contains("Timeframe", na=False) for col in df])
find_result = np.where(mask==True)
result = [find_result[0][0], find_result[1][0]]

Then output for df and result would be:

>>> df
   0  1                          2  3
0  1  2                          3  4
1  a  b  Timeframe for wave in____  d
2  5  6                          7  8
>>> result
[1, 2]

answered Jul 11, 2017 at 19:18

Cedric Zoppolo

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