CODEWITHSUNDEEP
pythonpandasnan
Alireza
Alireza

Reputation: 6848

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

This is my dataframe:

          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 Nan with []. How to do that? .fillna([]) did not work. I even tried replace(np.nan, []) but it gives error:

 TypeError('Invalid "to_replace" type: \'float\'',)

Upvotes: 93

Views: 83911

Answers (14)

Gian Arauz
Gian Arauz

Reputation: 456

Maybe not the most short/optimized solution, but I think is pretty readable:

# Masking-in nans
mask = df['ids'].isna()

# Filling nans with a list-like string and literally-evaluating such string
df.loc[mask, 'ids'] = df.loc[mask, 'ids'].fillna('[]').apply(eval)

EDIT

Following the advice from Swier's comment:

# Packages
import ast

# Masking-in nans
mask = df['ids'].isna()

# Filling nans with a list-like string and literally-evaluating such string
df.loc[mask, 'ids'] = df.loc[mask, 'ids'].fillna('[]').apply(ast.literal_eval)

Upvotes: 4

Yi Tang
Yi Tang

Reputation: 13

I have solved a more complex case and want to share the solution here.

![enter image description here

In each cell of the DataFrame there is a nested List with 100x sublists [a, b] inside. Some values for the columns bids_aggr3 and asks_aggr3 are np.nan. The number of NA columns is unknown. The example above shows only 2.

The goal is to use the pd.explode() to expand each row to 100x rows that containing one sublist of the original cell for each column respectively. This requires however that the values in each cell (nested lists) have same length. So i need to fill the NA's with a nested List like [[na, na], [na, na], ...[na, na]] with the length 100.

After some research i came to a generic solution which can replace NA's without specifying the columns.

nan_cell = [[np.nan, np.nan]]*100
rows = df.loc[df.isna().any(axis=1)].index
columns = df.columns[df.isna().any(axis=0)]
df.loc[rows, columns] = pd.Series([nan_cell]*len(rows))

In the line 1 the new nested List is generated. In the line 2 and 3 the rows and columns with NA's are located. In the line 4 use loc() to set the nested List as value for each NA cell. The loc() requires a Scala or Serie with the same length as input.

The result looks like this:

enter image description here

A quick check of the value in the first row:

enter image description here

HOWEVER! If the NA's are not occurring in sequence or not from the the top or the bottom of the column, rather somewhere mid in the dataframe and in different rows for each column, the solution above will not work.

In this case you can use the [index, column] pairs to directly access each cell to modify the values.

Example:

df = pd.DataFrame({'a': [0, 1, 2, np.nan, np.nan, 5, 6], 'b': [22, 23, 2, 1, 0, np.nan, 99]})

enter image description here

mask=df.isna().stack()
cells = mask.loc[mask].index.tolist()
cells
# idx, idy = np.where(pd.isnull(df))
# cells = np.column_stack([df.index[idx], df.columns[idy]])
# cells
for i in result:
    df.at[i[0], i[1]] = pd.Series([[np.nan, np,nan]])

enter image description here

The For loop could be performance bottleneck if the DataFrame is large. If someone knows a more pythonic/vectorized way, please share the solution.

Hopefully someone will find this helpful. Cheers!

Upvotes: 0

toto_tico
toto_tico

Reputation: 19037

Another solution that is explicit:

# use apply to only replace the nulls with the list  
df.loc[df.ids.isnull(), 'ids'] = df.loc[df.ids.isnull(), 'ids'].apply(lambda x: [])

Upvotes: 2

Иван Рычков
Иван Рычков

Reputation: 21

You can try this:

df.fillna(df.notna().applymap(lambda x: x or []))

Upvotes: 2

ronkov
ronkov

Reputation: 1583

A simple solution would be:

df['ids'].fillna("").apply(list)

As noted by @timgeb, this requires df['ids'] to contain lists or nan only.

Upvotes: 52

timgeb
timgeb

Reputation: 78700

Surprisingly, passing a dict with empty lists as values seems to work for Series.fillna, but not DataFrame.fillna - so if you want to work on a single column you can use this:

>>> df
     A    B    C
0  0.0  2.0  NaN
1  NaN  NaN  5.0
2  NaN  7.0  NaN
>>> df['C'].fillna({i: [] for i in df.index})
0    []
1     5
2    []
Name: C, dtype: object

The solution can be extended to DataFrames by applying it to every column.

>>> df.apply(lambda s: s.fillna({i: [] for i in df.index}))
    A   B   C
0   0   2  []
1  []  []   5
2  []   7  []

Note: for large Series/DataFrames with few missing values, this might create an unreasonable amount of throwaway empty lists.

Tested with pandas 1.0.5.

Upvotes: 13

botivegh
botivegh

Reputation: 500

This is probably faster, one liner solution:

df['ids'].fillna('DELETE').apply(lambda x : [] if x=='DELETE' else x)

Upvotes: 1

Allen Qin
Allen Qin

Reputation: 19947

Another solution using numpy:

df.ids = np.where(df.ids.isnull(), pd.Series([[]]*len(df)), df.ids)

Or using combine_first:

df.ids = df.ids.combine_first(pd.Series([[]]*len(df)))

Upvotes: 4

keramat
keramat

Reputation: 4543

Maybe more dense:

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

Upvotes: 1

TICH
TICH

Reputation: 17

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))

Upvotes: 0

Nick Edgar
Nick Edgar

Reputation: 1368

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.

Upvotes: 75

PlasmaBinturong
PlasmaBinturong

Reputation: 2274

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.

Upvotes: 41

hellpanderr
hellpanderr

Reputation: 5896

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))

Upvotes: 3

Alexander
Alexander

Reputation: 109546

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]

Upvotes: 26

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