我試圖将一列一分為二,但我知道我的資料中有空值。想象一下這個資料幀:df = pd.DataFrame(['fruit: apple','vegetable: asparagus',None, 'fruit: pear'], columns = ['text'])
df
text
0 fruit: apple
1 vegetable: asparagus
2 None
3 fruit: pear
我想把它分成多個列,如下所示:df['cat'] = df['text'].apply(lambda x: 'unknown' if x == None else x.split(': ')[0])
df['value'] = df['text'].apply(lambda x: 'unknown' if x == None else x.split(': ')[1])
print df
text cat value
0 fruit: apple fruit apple
1 vegetable: asparagus vegetable asparagus
2 None unknown unknown
3 fruit: pear fruit pear
但是,如果我有以下df:df = pd.DataFrame(['fruit: apple','vegetable: asparagus',np.nan, 'fruit: pear'], columns = ['text'])
拆分會導緻以下錯誤:df['cat'] = df['text'].apply(lambda x: 'unknown' if x == np.nan else x.split(': ')[0])
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
in ()
1 df = pd.DataFrame(['fruit: apple','vegetable: asparagus',np.nan, 'fruit: pear'], columns = ['text'])
2 #df.columns = ['col_name']
----> 3 df['cat'] = df['text'].apply(lambda x: 'unknown' if x == np.nan else x.split(': ')[0])
4 df['value'] = df['text'].apply(lambda x: 'unknown' if x == np.nan else x.split(': ')[1])
C:\Python27\lib\site-packages\pandas\core\series.pyc in apply(self, func, convert_dtype, args, **kwds)
2158 values = lib.map_infer(values, lib.Timestamp)
2159
-> 2160 mapped = lib.map_infer(values, f, convert=convert_dtype)
2161 if len(mapped) and isinstance(mapped[0], Series):
2162 from pandas.core.frame import DataFrame
pandas\src\inference.pyx in pandas.lib.map_infer (pandas\lib.c:62187)()
in (x)
1 df = pd.DataFrame(['fruit: apple','vegetable: asparagus',np.nan, 'fruit: pear'], columns = ['text'])
2 #df.columns = ['col_name']
----> 3 df['cat'] = df['text'].apply(lambda x: 'unknown' if x == np.nan else x.split(': ')[0])
4 df['value'] = df['text'].apply(lambda x: 'unknown' if x == np.nan else x.split(': ')[1])
AttributeError: 'float' object has no attribute 'split'
如何使用NaN值進行相同的拆分?
有沒有更好的方法來應用忽略空值的split函數?
假設這不是一個字元串示例,而是如果我有以下内容:df = pd.DataFrame([2,4,6,8,10,np.nan,12], columns = ['numerics'])
df['numerics'].apply(lambda x: np.nan if pd.isnull(x) else x/2.0)
我覺得Series.apply幾乎應該接受一個參數,該參數訓示它跳過空行并将它們作為空輸出。我還沒有找到一種更好的泛型方法來對序列進行轉換,而不必手動避免空值。