How to fill dataframe Nan values with empty list [] in pandas?

a 夏天 提交于 2019-12-02 20:27:17

You can first use loc to locate all rows that have a nan in the ids column, and then loop through these rows using at to set their values to an empty list:

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]

My approach is similar to @hellpanderrr's, but instead tests for list-ness rather than using 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.

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:

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 then enclose it in a list: the value you are assigning is a 2D array (1 column, isnull.sum() rows) containing empty lists as elements.

Without assignments:

1) Assuming we have only floats and integers in our dataframe

import math
df.apply(lambda x:x.apply(lambda x:[] if math.isnan(x) else x))

2) For any dataframe

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))

Maybe more dense:

df['ids'] = [[] if type(x) != list else x for x in df['ids']]
TICH

Create a function that checks your condition, if not, it returns an empty list/empty set etc.

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.

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))
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