CODEWITHSUNDEEP
pythonpandas
GreenGodot
GreenGodot

Reputation: 6753

Applying a function to a column that takes another column as an argument in Python Pandas

I want to apply a function to an entire column in a Pandas dataframe. This function will overwrite the data currently in that column but requires the value of another column next to it, to illustrate:

col 0, col 1,
 23,   'word'
 45,   'word2'
 63,   'word3'

I have tired passing in the number column into Pandas apply method:

df[1] = df.apply(retrieve_original_string(df[0]), axis=1)

But this throws an error:

sys:1: DtypeWarning: Columns (3,4) have mixed types. Specify dtype option on import or set low_memory=False.
Traceback (most recent call last):
  File "/home/noname365/similar_keywords_microsoft/similar_keywords.py", line 95, in <module>
    merged_df[1] = merged_df.apply(retrieve_original_string(merged_df[0], match_df), axis=1)
  File "/home/noname365/similar_keywords_microsoft/similar_keywords.py", line 12, in retrieve_original_string
    row_num = int(row)
  File "/home/noname365/virtualenvs/env35/lib/python3.5/site-packages/pandas/core/series.py", line 81, in wrapper
    "cannot convert the series to {0}".format(str(converter)))
TypeError: cannot convert the series to <class 'int'>

The error implies that I am passing the whole number column to the function instead of individually on a row-by-row basis. How would I accomplish this?

Upvotes: 3

Views: 328

Answers (1)

jezrael
jezrael

Reputation: 862441

IIUC you need iloc for selecting second column and add lambda as mentioned EdChum:

def retrieve_original_string(x):
    x = x + 4
    #add code
    return x


df.iloc[:,1] = df.apply(lambda x: retrieve_original_string(x[0]), axis=1)
print df
   col 0  col 1
0     23     27
1     45     49
2     63     67

#if you need new column
df['a'] = df.apply(lambda x: retrieve_original_string(x[0]), axis=1)
print df
   col 0    col 1   a
0     23   'word'  27
1     45  'word2'  49
2     63  'word3'  67

Or:

def retrieve_original_string(x):
    x = x + 4
    #add code
    return x


df.iloc[:,1] = df.iloc[:,0].apply(retrieve_original_string)
print df
   col 0  col 1
0     23     27
1     45     49
2     63     67

Upvotes: 2

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