【发布时间】:2017-03-29 22:32:55
【问题描述】:
我在我的数据框my_df 上使用apply,如下所示:
my_df['column_C'] = my_df.apply(lambda x : 'hello' if x['column_B'] is None else x['column_B'] )
我想要:
if x['column_B'] = None -> return 'hello'
if x['column_B'] != None -> return x['column_B']
然后我得到以下错误:
<ipython-input-31-aa087c9a635e> in <lambda>(x)
----> 1 my_df['column_C'] = my_df.apply(lambda x : 'hello' if x['column_B'] is None else x['column_B'] )
/usr/local/lib/python3.4/dist-packages/pandas/core/series.py in __getitem__(self, key)
599 key = com._apply_if_callable(key, self)
600 try:
--> 601 result = self.index.get_value(self, key)
602
603 if not is_scalar(result):
/usr/local/lib/python3.4/dist-packages/pandas/indexes/base.py in get_value(self, series, key)
2187 # python 3
2188 if is_scalar(key): # pragma: no cover
-> 2189 raise IndexError(key)
2190 raise InvalidIndexError(key)
2191
IndexError: ('column_B', 'occurred at index column_A')
有谁知道我在这里做错了什么?
【问题讨论】:
-
为什么不:
my_df['column_C'] = my_df['column_C'].fillna('hello')? -
这当然是一种更聪明的方法,谢谢!仍然对 apply 函数出了什么问题感到好奇......
标签: python-3.x pandas apply