In Python Pandas, what’s the best way to check whether a DataFrame has one (or more) NaN values?

I know about the function pd.isnan, but this returns a DataFrame of booleans for each element. This post right here doesn’t exactly answer my question either.

jwilner‘s response is spot on. I was exploring to see if there’s a faster option, since in my experience, summing flat arrays is (strangely) faster than counting. This code seems faster:

df.isnull().values.any()

For example:

In [2]: df = pd.DataFrame(np.random.randn(1000,1000))

In [3]: df[df > 0.9] = pd.np.nan

In [4]: %timeit df.isnull().any().any()
100 loops, best of 3: 14.7 ms per loop

In [5]: %timeit df.isnull().values.sum()
100 loops, best of 3: 2.15 ms per loop

In [6]: %timeit df.isnull().sum().sum()
100 loops, best of 3: 18 ms per loop

In [7]: %timeit df.isnull().values.any()
1000 loops, best of 3: 948 µs per loop

df.isnull().sum().sum() is a bit slower, but of course, has additional information — the number of NaNs.

You have a couple of options.

import pandas as pd
import numpy as np

df = pd.DataFrame(np.random.randn(10,6))
# Make a few areas have NaN values
df.iloc[1:3,1] = np.nan
df.iloc[5,3] = np.nan
df.iloc[7:9,5] = np.nan

Now the data frame looks something like this:

          0         1         2         3         4         5
0  0.520113  0.884000  1.260966 -0.236597  0.312972 -0.196281
1 -0.837552       NaN  0.143017  0.862355  0.346550  0.842952
2 -0.452595       NaN -0.420790  0.456215  1.203459  0.527425
3  0.317503 -0.917042  1.780938 -1.584102  0.432745  0.389797
4 -0.722852  1.704820 -0.113821 -1.466458  0.083002  0.011722
5 -0.622851 -0.251935 -1.498837       NaN  1.098323  0.273814
6  0.329585  0.075312 -0.690209 -3.807924  0.489317 -0.841368
7 -1.123433 -1.187496  1.868894 -2.046456 -0.949718       NaN
8  1.133880 -0.110447  0.050385 -1.158387  0.188222       NaN
9 -0.513741  1.196259  0.704537  0.982395 -0.585040 -1.693810

• Option 1: df.isnull().any().any() – This returns a boolean value

You know of the isnull() which would return a dataframe like this:

       0      1      2      3      4      5
0  False  False  False  False  False  False
1  False   True  False  False  False  False
2  False   True  False  False  False  False
3  False  False  False  False  False  False
4  False  False  False  False  False  False
5  False  False  False   True  False  False
6  False  False  False  False  False  False
7  False  False  False  False  False   True
8  False  False  False  False  False   True
9  False  False  False  False  False  False

If you make it df.isnull().any(), you can find just the columns that have NaN values:

0    False
1     True
2    False
3     True
4    False
5     True
dtype: bool

One more .any() will tell you if any of the above are True

> df.isnull().any().any()
True

• Option 2: df.isnull().sum().sum() – This returns an integer of the total number of NaN values:

This operates the same way as the .any().any() does, by first giving a summation of the number of NaN values in a column, then the summation of those values:

df.isnull().sum()
0    0
1    2
2    0
3    1
4    0
5    2
dtype: int64

Finally, to get the total number of NaN values in the DataFrame:

df.isnull().sum().sum()
5

To find out which rows have NaNs in a specific column:

nan_rows = df[df['name column'].isnull()]

If you need to know how many rows there are with “one or more NaNs”:

df.isnull().T.any().T.sum()

Or if you need to pull out these rows and examine them:

nan_rows = df[df.isnull().T.any().T]

df.isnull().any().any() should do it.

Adding to Hobs brilliant answer, I am very new to Python and Pandas so please point out if I am wrong.

To find out which rows have NaNs:

nan_rows = df[df.isnull().any(1)]

would perform the same operation without the need for transposing by specifying the axis of any() as 1 to check if ‘True’ is present in rows.

Super Simple Syntax: df.isna().any(axis=None)

Starting from v0.23.2, you can use DataFrame.isna + DataFrame.any(axis=None) where axis=None specifies logical reduction over the entire DataFrame.

# Setup
df = pd.DataFrame({'A': [1, 2, np.nan], 'B' : [np.nan, 4, 5]})
df
     A    B
0  1.0  NaN
1  2.0  4.0
2  NaN  5.0

df.isna()

       A      B
0  False   True
1  False  False
2   True  False

df.isna().any(axis=None)
# True

Useful Alternatives

numpy.isnan
Another performant option if you’re running older versions of pandas.

np.isnan(df.values)

array([[False,  True],
       [False, False],
       [ True, False]])

np.isnan(df.values).any()
# True

Alternatively, check the sum:

np.isnan(df.values).sum()
# 2

np.isnan(df.values).sum() > 0
# True

Series.hasnans
You can also iteratively call Series.hasnans. For example, to check if a single column has NaNs,

df['A'].hasnans
# True

And to check if any column has NaNs, you can use a comprehension with any (which is a short-circuiting operation).

any(df[c].hasnans for c in df)
# True

This is actually very fast.

Since none have mentioned, there is just another variable called hasnans.

df[i].hasnans will output to True if one or more of the values in the pandas Series is NaN, False if not. Note that its not a function.

pandas version ‘0.19.2’ and ‘0.20.2’

Since pandas has to find this out for DataFrame.dropna(), I took a look to see how they implement it and discovered that they made use of DataFrame.count(), which counts all non-null values in the DataFrame. Cf. pandas source code. I haven’t benchmarked this technique, but I figure the authors of the library are likely to have made a wise choice for how to do it.

let df be the name of the Pandas DataFrame and any value that is numpy.nan is a null value.

1. If you want to see which columns has nulls and which not(just True and False)

   df.isnull().any()
   

2. If you want to see only the columns that has nulls

   df.loc[:, df.isnull().any()].columns
   

3. If you want to see the count of nulls in every column

   df.isna().sum()
   

4. If you want to see the percentage of nulls in every column

   df.isna().sum()/(len(df))*100
   

5. If you want to see the percentage of nulls in columns only with nulls: df.loc[:,list(df.loc[:,df.isnull().any()].columns)].isnull().sum()/(len(df))*100

EDIT 1:

If you want to see where your data is missing visually:

import missingno
missingdata_df = df.columns[df.isnull().any()].tolist()
missingno.matrix(df[missingdata_df])

Just using math.isnan(x), Return True if x is a NaN (not a number), and False otherwise.

df.isnull().sum()

This will give you count of all NaN values present in the respective coloums of the DataFrame.

Here is another interesting way of finding null and replacing with a calculated value

    #Creating the DataFrame

    testdf = pd.DataFrame({'Tenure':[1,2,3,4,5],'Monthly':[10,20,30,40,50],'Yearly':[10,40,np.nan,np.nan,250]})
    >>> testdf2
       Monthly  Tenure  Yearly
    0       10       1    10.0
    1       20       2    40.0
    2       30       3     NaN
    3       40       4     NaN
    4       50       5   250.0

    #Identifying the rows with empty columns
    nan_rows = testdf2[testdf2['Yearly'].isnull()]
    >>> nan_rows
       Monthly  Tenure  Yearly
    2       30       3     NaN
    3       40       4     NaN

    #Getting the rows# into a list
    >>> index = list(nan_rows.index)
    >>> index
    [2, 3]

    # Replacing null values with calculated value
    >>> for i in index:
        testdf2['Yearly'][i] = testdf2['Monthly'][i] * testdf2['Tenure'][i]
    >>> testdf2
       Monthly  Tenure  Yearly
    0       10       1    10.0
    1       20       2    40.0
    2       30       3    90.0
    3       40       4   160.0
    4       50       5   250.0

I’ve been using the following and type casting it to a string and checking for the nan value

   (str(df.at[index, 'column']) == 'nan')

This allows me to check specific value in a series and not just return if this is contained somewhere within the series.

Or you can use .info() on the DF such as :

df.info(null_counts=True) which returns the number of non_null rows in a columns such as:

<class 'pandas.core.frame.DataFrame'>
Int64Index: 3276314 entries, 0 to 3276313
Data columns (total 10 columns):
n_matches                          3276314 non-null int64
avg_pic_distance                   3276314 non-null float64

The best would be to use:

df.isna().any().any()

Here is why. So isna() is used to define isnull(), but both of these are identical of course.

This is even faster than the accepted answer and covers all 2D panda arrays.

import missingno as msno
msno.matrix(df)  # just to visualize. no missing value.

df.apply(axis=0, func=lambda x : any(pd.isnull(x)))

Will check for each column if it contains Nan or not.

We can see the null values present in the dataset by generating heatmap using seaborn moduleheatmap

import pandas as pd
import seaborn as sns
dataset=pd.read_csv('train.csv')
sns.heatmap(dataset.isnull(),cbar=False)

You could not only check if any ‘NaN’ exist but also get the percentage of ‘NaN’s in each column using the following,

df = pd.DataFrame({'col1':[1,2,3,4,5],'col2':[6,np.nan,8,9,10]})  
df  

   col1 col2  
0   1   6.0  
1   2   NaN  
2   3   8.0  
3   4   9.0  
4   5   10.0  


df.isnull().sum()/len(df)  
col1    0.0  
col2    0.2  
dtype: float64

Depending on the type of data you’re dealing with, you could also just get the value counts of each column while performing your EDA by setting dropna to False.

for col in df:
   print df[col].value_counts(dropna=False)

Works well for categorical variables, not so much when you have many unique values.