pandas 如何在熊猫中用空列表[]填充数据框Nan值？
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How to fill dataframe Nan values with empty list [] in pandas?
提问by ALH
This is my dataframe:
这是我的数据框：
date ids
0 2011-04-23 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,...
1 2011-04-24 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,...
2 2011-04-25 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,...
3 2011-04-26 Nan
4 2011-04-27 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,...
5 2011-04-28 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,...
I want to replace Nanwith []. How to do that? Fillna([]) did not work. I even tried replace(np.nan, [])but it gives error:
我想Nan用[]替换。怎么做？Fillna([]) 不起作用。我什至尝试过，replace(np.nan, [])但它给出了错误：
TypeError('Invalid "to_replace" type: \'float\'',)
采纳答案by Alexander
You can first use locto locate all rows that have a nanin the idscolumn, and then loop through these rows using atto set their values to an empty list:
你可以先使用loc以找出有所有行nan的ids列，然后通过使用这些行循环at到它们的值设置为空列表：
for row in df.loc[df.ids.isnull(), 'ids'].index:
df.at[row, 'ids'] = []
>>> df
date ids
0 2011-04-23 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
1 2011-04-24 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
2 2011-04-25 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
3 2011-04-26 []
4 2011-04-27 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
5 2011-04-28 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
回答by Nick Edgar
My approach is similar to @hellpanderrr's, but instead tests for list-ness rather than using isnan:
我的方法类似于@hellpanderrr 的方法，但是测试列表性而不是使用isnan：
df['ids'] = df['ids'].apply(lambda d: d if isinstance(d, list) else [])
I originally tried using pd.isnull(or pd.notnull) but, when given a list, that returns the null-ness of each element.
我最初尝试使用pd.isnull(or pd.notnull) 但是，当给定一个列表时，它返回每个元素的空性。
回答by PlasmaBinturong
After a lot of head-scratching I found this method that should be the most efficient (no looping, no apply), just assigning to a slice:
经过大量的挠头后，我发现这种方法应该是最有效的（没有循环，没有应用），只需分配给一个切片：
isnull = df.ids.isnull()
df.loc[isnull, 'ids'] = [ [[]] * isnull.sum() ]
The trick was to construct your list of []of the right size (isnull.sum()), and thenenclose it in a list: the value you are assigning is a 2Darray (1 column, isnull.sum()rows) containing empty lists as elements.
诀窍是构建[]正确大小 ( isnull.sum())的列表，然后将其包含在一个列表中：您分配的值是一个包含空列表作为元素的二维数组（1 列，isnull.sum()行）。
回答by hellpanderr
Without assignments:
没有任务：
1) Assuming we have only floats and integers in our dataframe
1）假设我们的数据框中只有浮点数和整数
import math
df.apply(lambda x:x.apply(lambda x:[] if math.isnan(x) else x))
2) For any dataframe
2）对于任何数据帧
import math
def isnan(x):
if isinstance(x, (int, long, float, complex)) and math.isnan(x):
return True
df.apply(lambda x:x.apply(lambda x:[] if isnan(x) else x))
回答by Allen
Another solution using numpy:
使用 numpy 的另一种解决方案：
df.ids = np.where(df.ids.isnull(), pd.Series([[]]*len(df)), df.ids)
Or using combine_first:
或者使用 combine_first：
df.ids = df.ids.combine_first(pd.Series([[]]*len(df)))
回答by botivegh
This is probably faster, one liner solution:
这可能更快，一种班轮解决方案：
df['ids'].fillna('DELETE').apply(lambda x : [] if x=='DELETE' else x)
回答by keramat
Maybe more dense:
也许更密集：
df['ids'] = [[] if type(x) != list else x for x in df['ids']]
回答by TICH
Create a function that checks your condition, if not, it returns an empty list/empty set etc.
创建一个函数来检查你的条件，如果没有，它返回一个空列表/空集等。
Then apply that function to the variable, but also assigning the new calculated variable to the old one or to a new variable if you wish.
然后将该函数应用于变量，但也可以根据需要将新计算的变量分配给旧变量或新变量。
aa=pd.DataFrame({'d':[1,1,2,3,3,np.NaN],'r':[3,5,5,5,5,'e']})
def check_condition(x):
if x>0:
return x
else:
return list()
aa['d]=aa.d.apply(lambda x:check_condition(x))
