I want to find the number of NaN
in each column of my data.
df.info()
does not return a DataFame, the method only prints the information.
df.info()
will give the data types and non-null counts for each column
Use the isna()
method (or it's alias isnull()
which is also compatible with older pandas versions < 0.21.0) and then sum to count the NaN values. For one column:
>>> s = pd.Series([1,2,3, np.nan, np.nan])
>>> s.isna().sum() # or s.isnull().sum() for older pandas versions
2
For several columns, this also works:
>>> df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
>>> df.isna().sum()
a 1
b 2
dtype: int64
Lets assume df
is a pandas DataFrame.
Then,
df.isnull().sum(axis = 0)
This will give number of NaN values in every column.
If you need, NaN values in every row,
df.isnull().sum(axis = 1)
You could subtract the total length from the count of non-nan values:
count_nan = len(df) - df.count()
You should time it on your data. For small Series got a 3x speed up in comparison with the isnull
solution.
isnull
is already two times faster as this.
Based on the most voted answer we can easily define a function that gives us a dataframe to preview the missing values and the % of missing values in each column:
def missing_values_table(df):
mis_val = df.isnull().sum()
mis_val_percent = 100 * df.isnull().sum() / len(df)
mis_val_table = pd.concat([mis_val, mis_val_percent], axis=1)
mis_val_table_ren_columns = mis_val_table.rename(
columns = {0 : 'Missing Values', 1 : '% of Total Values'})
mis_val_table_ren_columns = mis_val_table_ren_columns[
mis_val_table_ren_columns.iloc[:,1] != 0].sort_values(
'% of Total Values', ascending=False).round(1)
print ("Your selected dataframe has " + str(df.shape[1]) + " columns.\n"
"There are " + str(mis_val_table_ren_columns.shape[0]) +
" columns that have missing values.")
return mis_val_table_ren_columns
Since pandas 0.14.1 my suggestion here to have a keyword argument in the value_counts method has been implemented:
import pandas as pd
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
for col in df:
print df[col].value_counts(dropna=False)
2 1
1 1
NaN 1
dtype: int64
NaN 2
1 1
dtype: int64
if its just counting nan values in a pandas column here is a quick way
import pandas as pd
## df1 as an example data frame
## col1 name of column for which you want to calculate the nan values
sum(pd.isnull(df1['col1']))
The below will print all the Nan columns in descending order.
df.isnull().sum().sort_values(ascending = False)
or
The below will print first 15 Nan columns in descending order.
df.isnull().sum().sort_values(ascending = False).head(15)
df.isnull().sum()
will give the column-wise sum of missing values.
If you want to know the sum of missing values in a particular column then following code will work: df.column.isnull().sum()
df.isnull().sum()
//type: <class 'pandas.core.series.Series'>
or
df.column_name.isnull().sum()
//type: <type 'numpy.int64'>
if you are using Jupyter Notebook, How about....
%%timeit
df.isnull().any().any()
or
%timeit
df.isnull().values.sum()
or, are there anywhere NaNs in the data, if yes, where?
df.isnull().any()
import numpy as np
import pandas as pd
raw_data = {'first_name': ['Jason', np.nan, 'Tina', 'Jake', 'Amy'],
'last_name': ['Miller', np.nan, np.nan, 'Milner', 'Cooze'],
'age': [22, np.nan, 23, 24, 25],
'sex': ['m', np.nan, 'f', 'm', 'f'],
'Test1_Score': [4, np.nan, 0, 0, 0],
'Test2_Score': [25, np.nan, np.nan, 0, 0]}
results = pd.DataFrame(raw_data, columns = ['first_name', 'last_name', 'age', 'sex', 'Test1_Score', 'Test2_Score'])
results
'''
first_name last_name age sex Test1_Score Test2_Score
0 Jason Miller 22.0 m 4.0 25.0
1 NaN NaN NaN NaN NaN NaN
2 Tina NaN 23.0 f 0.0 NaN
3 Jake Milner 24.0 m 0.0 0.0
4 Amy Cooze 25.0 f 0.0 0.0
'''
You can use following function, which will give you output in Dataframe
Zero Values
Missing Values
% of Total Values
Total Zero Missing Values
% Total Zero Missing Values
Data Type
Just copy and paste following function and call it by passing your pandas Dataframe
def missing_zero_values_table(df):
zero_val = (df == 0.00).astype(int).sum(axis=0)
mis_val = df.isnull().sum()
mis_val_percent = 100 * df.isnull().sum() / len(df)
mz_table = pd.concat([zero_val, mis_val, mis_val_percent], axis=1)
mz_table = mz_table.rename(
columns = {0 : 'Zero Values', 1 : 'Missing Values', 2 : '% of Total Values'})
mz_table['Total Zero Missing Values'] = mz_table['Zero Values'] + mz_table['Missing Values']
mz_table['% Total Zero Missing Values'] = 100 * mz_table['Total Zero Missing Values'] / len(df)
mz_table['Data Type'] = df.dtypes
mz_table = mz_table[
mz_table.iloc[:,1] != 0].sort_values(
'% of Total Values', ascending=False).round(1)
print ("Your selected dataframe has " + str(df.shape[1]) + " columns and " + str(df.shape[0]) + " Rows.\n"
"There are " + str(mz_table.shape[0]) +
" columns that have missing values.")
# mz_table.to_excel('D:/sampledata/missing_and_zero_values.xlsx', freeze_panes=(1,0), index = False)
return mz_table
missing_zero_values_table(results)
Output
Your selected dataframe has 6 columns and 5 Rows.
There are 6 columns that have missing values.
Zero Values Missing Values % of Total Values Total Zero Missing Values % Total Zero Missing Values Data Type
last_name 0 2 40.0 2 40.0 object
Test2_Score 2 2 40.0 4 80.0 float64
first_name 0 1 20.0 1 20.0 object
age 0 1 20.0 1 20.0 float64
sex 0 1 20.0 1 20.0 object
Test1_Score 3 1 20.0 4 80.0 float64
If you want to keep it simple then you can use following function to get missing values in %
def missing(dff):
print (round((dff.isnull().sum() * 100/ len(dff)),2).sort_values(ascending=False))
missing(results)
'''
Test2_Score 40.0
last_name 40.0
Test1_Score 20.0
sex 20.0
age 20.0
first_name 20.0
dtype: float64
'''
Please use below for particular column count
dataframe.columnName.isnull().sum()
To count zeroes:
df[df == 0].count(axis=0)
To count NaN:
df.isnull().sum()
or
df.isna().sum()
Hope this helps,
import pandas as pd
import numpy as np
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan],'c':[np.nan,2,np.nan], 'd':[np.nan,np.nan,np.nan]})
https://i.stack.imgur.com/f32tg.png
df.isnull().sum()/len(df) * 100
https://i.stack.imgur.com/jYryw.png
Thres = 40
(df.isnull().sum()/len(df) * 100 ) < Thres
https://i.stack.imgur.com/C3X3t.png
You can use value_counts method and print values of np.nan
s.value_counts(dropna = False)[np.nan]
s.value_counts(dropna = False)
One other simple option not suggested yet, to just count NaNs, would be adding in the shape to return the number of rows with NaN.
df[df['col_name'].isnull()]['col_name'].shape
For the 1st part count NaN
we have multiple way.
Method 1 count
, due to the count
will ignore the NaN
which is different from size
print(len(df) - df.count())
Method 2 isnull
/ isna
chain with sum
print(df.isnull().sum())
#print(df.isna().sum())
Method 3 describe
/ info
: notice this will output the 'notnull' value count
print(df.describe())
#print(df.info())
Method from numpy
print(np.count_nonzero(np.isnan(df.values),axis=0))
For the 2nd part of the question, If we would like drop the column by the thresh,we can try with dropna
thresh, optional Require that many non-NA values.
Thresh = n # no null value require, you can also get the by int(x% * len(df))
df = df.dropna(thresh = Thresh, axis = 1)
df1.isnull().sum()
This will do the trick.
Here is the code for counting Null
values column wise :
df.isna().sum()
There is a nice Dzone article from July 2017 which details various ways of summarising NaN values. Check it out here.
The article I have cited provides additional value by: (1) Showing a way to count and display NaN counts for every column so that one can easily decide whether or not to discard those columns and (2) Demonstrating a way to select those rows in specific which have NaNs so that they may be selectively discarded or imputed.
Here's a quick example to demonstrate the utility of the approach - with only a few columns perhaps its usefulness is not obvious but I found it to be of help for larger data-frames.
import pandas as pd
import numpy as np
# example DataFrame
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
# Check whether there are null values in columns
null_columns = df.columns[df.isnull().any()]
print(df[null_columns].isnull().sum())
# One can follow along further per the cited article
In case you need to get the non-NA (non-None) and NA (None) counts across different groups pulled out by groupby:
gdf = df.groupby(['ColumnToGroupBy'])
def countna(x):
return (x.isna()).sum()
gdf.agg(['count', countna, 'size'])
This returns the counts of non-NA, NA and total number of entries per group.
You can try with:
In [1]: s = pd.DataFrame('a'=[1,2,5, np.nan, np.nan,3],'b'=[1,3, np.nan, np.nan,3,np.nan])
In [4]: s.isna().sum()
Out[4]: out = {'a'=2, 'b'=3} # the number of NaN values for each column
If needed the gran total of nans:
In [5]: s.isna().sum().sum()
Out[6]: out = 5 #the inline sum of Out[4]
based to the answer that was given and some improvements this is my approach
def PercentageMissin(Dataset):
"""this function will return the percentage of missing values in a dataset """
if isinstance(Dataset,pd.DataFrame):
adict={} #a dictionary conatin keys columns names and values percentage of missin value in the columns
for col in Dataset.columns:
adict[col]=(np.count_nonzero(Dataset[col].isnull())*100)/len(Dataset[col])
return pd.DataFrame(adict,index=['% of missing'],columns=adict.keys())
else:
raise TypeError("can only be used with panda dataframe")
df.apply(lambda x: x.value_counts(dropna=False)[np.nan]/x.size*100)
I use this loop to count missing values for each column:
# check missing values
import numpy as np, pandas as pd
for col in df:
print(col +': '+ np.str(df[col].isna().sum()))
You can use df.iteritems() to loop over the data frame. Set a conditional within a for loop to calculate the NaN values percent for each column, and drop those that contain a value of NaNs over your set threshold:
for col, val in df.iteritems():
if (df[col].isnull().sum() / len(val) * 100) > 30:
df.drop(columns=col, inplace=True)
Used the solution proposed by @sushmit in my code.
A possible variation of the same can also be
colNullCnt = []
for z in range(len(df1.cols)):
colNullCnt.append([df1.cols[z], sum(pd.isnull(trainPd[df1.cols[z]]))])
Advantage of this is that it returns the result for each of the columns in the df henceforth.
import pandas as pd
import numpy as np
# example DataFrame
df = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
# count the NaNs in a column
num_nan_a = df.loc[ (pd.isna(df['a'])) , 'a' ].shape[0]
num_nan_b = df.loc[ (pd.isna(df['b'])) , 'b' ].shape[0]
# summarize the num_nan_b
print(df)
print(' ')
print(f"There are {num_nan_a} NaNs in column a")
print(f"There are {num_nan_b} NaNs in column b")
Gives as output:
a b
0 1.0 NaN
1 2.0 1.0
2 NaN NaN
There are 1 NaNs in column a
There are 2 NaNs in column b
Suppose you want to get the number of missing values(NaN) in a column(series) known as price in a dataframe called reviews
#import the dataframe
import pandas as pd
reviews = pd.read_csv("../input/wine-reviews/winemag-data-130k-v2.csv", index_col=0)
To get the missing values, with n_missing_prices as the variable, simple do
n_missing_prices = sum(reviews.price.isnull())
print(n_missing_prices)
sum is the key method here, was trying to use count before i realized sum is the right method to use in this context
I've written a short function (Python 3) to produce .info as a pandas dataframe that can be then be written to excel:
df1 = pd.DataFrame({'a':[1,2,np.nan], 'b':[np.nan,1,np.nan]})
def info_as_df (df):
null_counts = df.isna().sum()
info_df = pd.DataFrame(list(zip(null_counts.index,null_counts.values))\
, columns = ['Column', 'Nulls_Count'])
data_types = df.dtypes
info_df['Dtype'] = data_types.values
return info_df
print(df1.info())
print(info_as_df(df1))
Which gives:
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 3 entries, 0 to 2
Data columns (total 2 columns):
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 a 2 non-null float64
1 b 1 non-null float64
dtypes: float64(2)
memory usage: 176.0 bytes
None
Column Nulls_Count Dtype
0 a 1 float64
1 b 2 float64
Another way just for completeness is using np.count_nonzero
with .isna():
np.count_nonzero(df.isna())
%timeit np.count_nonzero(df.isna())
512 ms ± 24.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Comparing with the top answers using 1000005 rows × 16 columns dataframe:
%timeit df.isna().sum()
492 ms ± 55.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit df.isnull().sum(axis = 0)
478 ms ± 34.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit count_nan = len(df) - df.count()
484 ms ± 47.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
data:
raw_data = {'first_name': ['Jason', np.nan, 'Tina', 'Jake', 'Amy'],
'last_name': ['Miller', np.nan, np.nan, 'Milner', 'Cooze'],
'age': [22, np.nan, 23, 24, 25],
'sex': ['m', np.nan, 'f', 'm', 'f'],
'Test1_Score': [4, np.nan, 0, 0, 0],
'Test2_Score': [25, np.nan, np.nan, 0, 0]}
results = pd.DataFrame(raw_data, columns = ['first_name', 'last_name', 'age', 'sex', 'Test1_Score', 'Test2_Score'])
# big dataframe for %timeit
big_df = pd.DataFrame(np.random.randint(0, 100, size=(1000000, 10)), columns=list('ABCDEFGHIJ'))
df = pd.concat([big_df,results]) # 1000005 rows × 16 columns
Success story sharing
df
you can usedf.isnull().sum().sum()
.sum(axis=0)
, which is the default behavior. And to get rowsums,.sum(axis=1)
.df.isnull().values.sum()
df['column_name'].isna().sum()
also works if anyone is wondering.df.isna()
produces Boolean Series where the number ofTrue
is the number ofNaN
, anddf.isna().sum()
addsFalse
andTrue
replacing them respectively by 0 and 1. Therefore this indirectly counts theNaN
, where a simplecount
would just return the length of the column.