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.
If you need to know how many rows there are with "one or more NaN
s":
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()]
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.
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.
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 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
df.isnull().any().any()
- This returns a boolean valueYou 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
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
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
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 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.
df.apply(axis=0, func=lambda x : any(pd.isnull(x)))
Will check for each column if it contains Nan or not.
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
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.
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'
df.isnull().any().any()
should do it.
let df
be the name of the Pandas DataFrame and any value that is numpy.nan
is a null value.
If you want to see which columns has nulls and which do not(just True and False)
df.isnull().any()
If you want to see only the columns that has nulls
df.loc[:, df.isnull().any()].columns
If you want to see the count of nulls in every column
df.isna().sum()
If you want to see the percentage of nulls in every column
df.isna().sum()/(len(df))*100
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])
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
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
df.isnull().sum()
This will give you count of all NaN values present in the respective coloums of the DataFrame.
To find out which rows have NaNs in a specific column:
nan_rows = df[df['name column'].isnull()]
Just using math.isnan(x), Return True if x is a NaN (not a number), and False otherwise.
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.
Source: Stackoverflow.com