将函数应用于将另一列作为Python Pandas中的参数的列

时间:2016-03-30 10:55:43

标签: python pandas

我想将函数应用于Pandas数据帧中的整个列。此函数将覆盖该列中当前的数据,但需要其旁边的另一列的值,以说明:

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

我已经厌倦了将数字列传入Pandas应用方法:

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

但这会引发错误:

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'>

错误意味着我将整个数列传递给函数,而不是逐行传递。我该如何做到这一点?

1 个答案:

答案 0 :(得分:2)

IIUC您需要iloc来选择第二列并按照EdChum添加lambda

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

或者:

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