pandas DataFrame: replace nan values with average of columns

PythonPandasNan

Python Problem Overview


I've got a pandas DataFrame filled mostly with real numbers, but there is a few nan values in it as well.

How can I replace the nans with averages of columns where they are?

This question is very similar to this one: https://stackoverflow.com/questions/18689235/numpy-array-replace-nan-values-with-average-of-columns but, unfortunately, the solution given there doesn't work for a pandas DataFrame.

Python Solutions


Solution 1 - Python

You can simply use DataFrame.fillna to fill the nan's directly:

In [27]: df 
Out[27]: 
          A         B         C
0 -0.166919  0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3       NaN -2.027325  1.533582
4       NaN       NaN  0.461821
5 -0.788073       NaN       NaN
6 -0.916080 -0.612343       NaN
7 -0.887858  1.033826       NaN
8  1.948430  1.025011 -2.982224
9  0.019698 -0.795876 -0.046431

In [28]: df.mean()
Out[28]: 
A   -0.151121
B   -0.231291
C   -0.530307
dtype: float64

In [29]: df.fillna(df.mean())
Out[29]: 
          A         B         C
0 -0.166919  0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 -0.151121 -2.027325  1.533582
4 -0.151121 -0.231291  0.461821
5 -0.788073 -0.231291 -0.530307
6 -0.916080 -0.612343 -0.530307
7 -0.887858  1.033826 -0.530307
8  1.948430  1.025011 -2.982224
9  0.019698 -0.795876 -0.046431

The docstring of fillna says that value should be a scalar or a dict, however, it seems to work with a Series as well. If you want to pass a dict, you could use df.mean().to_dict().

Solution 2 - Python

Try:

sub2['income'].fillna((sub2['income'].mean()), inplace=True)

Solution 3 - Python

In [16]: df = DataFrame(np.random.randn(10,3))

In [17]: df.iloc[3:5,0] = np.nan

In [18]: df.iloc[4:6,1] = np.nan

In [19]: df.iloc[5:8,2] = np.nan

In [20]: df
Out[20]: 
          0         1         2
0  1.148272  0.227366 -2.368136
1 -0.820823  1.071471 -0.784713
2  0.157913  0.602857  0.665034
3       NaN -0.985188 -0.324136
4       NaN       NaN  0.238512
5  0.769657       NaN       NaN
6  0.141951  0.326064       NaN
7 -1.694475 -0.523440       NaN
8  0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794

In [22]: df.mean()
Out[22]: 
0   -0.251534
1   -0.040622
2   -0.841219
dtype: float64

Apply per-column the mean of that columns and fill

In [23]: df.apply(lambda x: x.fillna(x.mean()),axis=0)
Out[23]: 
          0         1         2
0  1.148272  0.227366 -2.368136
1 -0.820823  1.071471 -0.784713
2  0.157913  0.602857  0.665034
3 -0.251534 -0.985188 -0.324136
4 -0.251534 -0.040622  0.238512
5  0.769657 -0.040622 -0.841219
6  0.141951  0.326064 -0.841219
7 -1.694475 -0.523440 -0.841219
8  0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794

Solution 4 - Python

Although, the below code does the job, BUT its performance takes a big hit, as you deal with a DataFrame with # records 100k or more:

df.fillna(df.mean())

In my experience, one should replace NaN values (be it with Mean or Median), only where it is required, rather than applying fillna() all over the DataFrame.

I had a DataFrame with 20 variables, and only 4 of them required NaN values treatment (replacement). I tried the above code (Code 1), along with a slightly modified version of it (code 2), where i ran it selectively .i.e. only on variables which had a NaN value

#------------------------------------------------
#----(Code 1) Treatment on overall DataFrame-----

df.fillna(df.mean())

#------------------------------------------------
#----(Code 2) Selective Treatment----------------

for i in df.columns[df.isnull().any(axis=0)]:     #---Applying Only on variables with NaN values
    df[i].fillna(df[i].mean(),inplace=True)

#---df.isnull().any(axis=0) gives True/False flag (Boolean value series), 
#---which when applied on df.columns[], helps identify variables with NaN values

Below is the performance i observed, as i kept on increasing the # records in DataFrame

DataFrame with ~100k records

  • Code 1: 22.06 Seconds
  • Code 2: 0.03 Seconds

DataFrame with ~200k records

  • Code 1: 180.06 Seconds
  • Code 2: 0.06 Seconds

DataFrame with ~1.6 Million records

  • Code 1: code kept running endlessly
  • Code 2: 0.40 Seconds

DataFrame with ~13 Million records

  • Code 1: --did not even try, after seeing performance on 1.6 Mn records--
  • Code 2: 3.20 Seconds

Apologies for a long answer ! Hope this helps !

Solution 5 - Python

If you want to impute missing values with mean and you want to go column by column, then this will only impute with the mean of that column. This might be a little more readable.

sub2['income'] = sub2['income'].fillna((sub2['income'].mean()))

Solution 6 - Python

# To read data from csv file
Dataset = pd.read_csv('Data.csv')

X = Dataset.iloc[:, :-1].values

# To calculate mean use imputer class
from sklearn.impute import SimpleImputer
imputer = SimpleImputer(missing_values=np.nan, strategy='mean')
imputer = imputer.fit(X[:, 1:3])
X[:, 1:3] = imputer.transform(X[:, 1:3])

Solution 7 - Python

Directly use df.fillna(df.mean()) to fill all the null value with mean

If you want to fill null value with mean of that column then you can use this

suppose x=df['Item_Weight'] here Item_Weight is column name

here we are assigning (fill null values of x with mean of x into x)

df['Item_Weight'] = df['Item_Weight'].fillna((df['Item_Weight'].mean()))

If you want to fill null value with some string then use

here Outlet_size is column name

df.Outlet_Size = df.Outlet_Size.fillna('Missing')

Solution 8 - Python

Pandas: How to replace NaN (nan) values with the average (mean), median or other statistics of one column

Say your DataFrame is df and you have one column called nr_items. This is: df['nr_items']

If you want to replace the NaN values of your column df['nr_items'] with the mean of the column:

Use method .fillna():

mean_value=df['nr_items'].mean() df['nr_item_ave']=df['nr_items'].fillna(mean_value)

I have created a new df column called nr_item_ave to store the new column with the NaN values replaced by the mean value of the column.

You should be careful when using the mean. If you have outliers is more recommendable to use the median

Solution 9 - Python

Another option besides those above is:

df = df.groupby(df.columns, axis = 1).transform(lambda x: x.fillna(x.mean()))

It's less elegant than previous responses for mean, but it could be shorter if you desire to replace nulls by some other column function.

Solution 10 - Python

using sklearn library preprocessing class

from sklearn.impute import SimpleImputer
missingvalues = SimpleImputer(missing_values = np.nan, strategy = 'mean', axis = 0)
missingvalues = missingvalues.fit(x[:,1:3])
x[:,1:3] = missingvalues.transform(x[:,1:3])

Note: In the recent version parameter missing_values value change to np.nan from NaN

Solution 11 - Python

I use this method to fill missing values by average of a column.

fill_mean = lambda col : col.fillna(col.mean())

df = df.apply(fill_mean, axis = 0)

Solution 12 - Python

You can also use value_counts to get the most frequent values. This would work on different datatypes.

df = df.apply(lambda x:x.fillna(x.value_counts().index[0]))

Here is the value_counts api reference.

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