44

I currently came up with some work arounds to count the number of missing values in a pandas DataFrame. Those are quite ugly and I am wondering if there is a better way to do it.

Let's create an example DataFrame:

from numpy.random import randn
df = pd.DataFrame(randn(5, 3), index=['a', 'c', 'e', 'f', 'h'],
               columns=['one', 'two', 'three'])
df = df.reindex(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])

enter image description here

What I currently have is

a) Counting cells with missing values:

>>> sum(df.isnull().values.ravel())
9

b) Counting rows that have missing values somewhere:

>>> sum([True for idx,row in df.iterrows() if any(row.isnull())])
3
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    Regarding counting the number of rows that have somewhere missing values: `len(df) - len(df.dropna())` – NeStack Sep 18 '21 at 18:25

9 Answers9

47

For the second count I think just subtract the number of rows from the number of rows returned from dropna:

In [14]:

from numpy.random import randn
df = pd.DataFrame(randn(5, 3), index=['a', 'c', 'e', 'f', 'h'],
               columns=['one', 'two', 'three'])
df = df.reindex(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h'])
df
Out[14]:
        one       two     three
a -0.209453 -0.881878  3.146375
b       NaN       NaN       NaN
c  0.049383 -0.698410 -0.482013
d       NaN       NaN       NaN
e -0.140198 -1.285411  0.547451
f -0.219877  0.022055 -2.116037
g       NaN       NaN       NaN
h -0.224695 -0.025628 -0.703680
In [18]:

df.shape[0] - df.dropna().shape[0]
Out[18]:
3

The first could be achieved using the built in methods:

In [30]:

df.isnull().values.ravel().sum()
Out[30]:
9

Timings

In [34]:

%timeit sum([True for idx,row in df.iterrows() if any(row.isnull())])
%timeit df.shape[0] - df.dropna().shape[0]
%timeit sum(map(any, df.apply(pd.isnull)))
1000 loops, best of 3: 1.55 ms per loop
1000 loops, best of 3: 1.11 ms per loop
1000 loops, best of 3: 1.82 ms per loop
In [33]:

%timeit sum(df.isnull().values.ravel())
%timeit df.isnull().values.ravel().sum()
%timeit df.isnull().sum().sum()
1000 loops, best of 3: 215 µs per loop
1000 loops, best of 3: 210 µs per loop
1000 loops, best of 3: 605 µs per loop

So my alternatives are a little faster for a df of this size

Update

So for a df with 80,000 rows I get the following:

In [39]:

%timeit sum([True for idx,row in df.iterrows() if any(row.isnull())])
%timeit df.shape[0] - df.dropna().shape[0]
%timeit sum(map(any, df.apply(pd.isnull)))
%timeit np.count_nonzero(df.isnull())
1 loops, best of 3: 9.33 s per loop
100 loops, best of 3: 6.61 ms per loop
100 loops, best of 3: 3.84 ms per loop
1000 loops, best of 3: 395 µs per loop
In [40]:

%timeit sum(df.isnull().values.ravel())
%timeit df.isnull().values.ravel().sum()
%timeit df.isnull().sum().sum()
%timeit np.count_nonzero(df.isnull().values.ravel())
1000 loops, best of 3: 675 µs per loop
1000 loops, best of 3: 679 µs per loop
100 loops, best of 3: 6.56 ms per loop
1000 loops, best of 3: 368 µs per loop

Actually np.count_nonzero wins this hands down.

EdChum
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    I think your answer for counting the number of rows that have somewhere missing values can be stripped of `shape` and one can just do `len(df) - len(df.dropna())` – NeStack Sep 18 '21 at 18:18
17

So many wrong answers here. OP asked for number of rows with null values, not columns.

Here is a better example:

from numpy.random import randn
df = pd.DataFrame(randn(5, 3), index=['a', 'c', 'e', 'f', 'h'],columns=['one','two', 'three'])
df = df.reindex(['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h','asdf'])
print(df)

`Now there is obviously 4 rows with null values.

           one       two     three
a    -0.571617  0.952227  0.030825
b          NaN       NaN       NaN
c     0.627611 -0.462141  1.047515
d          NaN       NaN       NaN
e     0.043763  1.351700  1.480442
f     0.630803  0.931862  1.500602
g          NaN       NaN       NaN
h     0.729103 -1.198237 -0.207602
asdf       NaN       NaN       NaN

You would get answer as 3 (number of columns with NaNs) if you used some of the answers here. Fuentes' answer works.

Here is how I got it:

df.isnull().any(axis=1).sum()
#4
timeit df.isnull().any(axis=1).sum()
#10000 loops, best of 3: 193 µs per loop

'Fuentes':

sum(df.apply(lambda x: sum(x.isnull().values), axis = 1)>0)
#4
timeit sum(df.apply(lambda x: sum(x.isnull().values), axis = 1)>0)
#1000 loops, best of 3: 677 µs per loop
ConanG
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12

What about numpy.count_nonzero:

 np.count_nonzero(df.isnull().values)   
 np.count_nonzero(df.isnull())           # also works

count_nonzero is pretty quick. However, I constructed a dataframe from a (1000,1000) array and randomly inserted 100 nan values at different positions and measured the times of the various answers in iPython:

%timeit np.count_nonzero(df.isnull().values)
1000 loops, best of 3: 1.89 ms per loop

%timeit df.isnull().values.ravel().sum()
100 loops, best of 3: 3.15 ms per loop

%timeit df.isnull().sum().sum()
100 loops, best of 3: 15.7 ms per loop

Not a huge time improvement over the OPs original but possibly less confusing in the code, your decision. There isn't really any difference in execution time between the two count_nonzero methods (with and without .values).

Paul Joireman
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  • Documentation link for convenience: https://docs.scipy.org/doc/numpy/reference/generated/numpy.count_nonzero.html – Cole Murray Nov 25 '19 at 06:18
11

A simple approach to counting the missing values in the rows or in the columns

df.apply(lambda x: sum(x.isnull().values), axis = 0) # For columns
df.apply(lambda x: sum(x.isnull().values), axis = 1) # For rows

Number of rows with at least one missing value:

sum(df.apply(lambda x: sum(x.isnull().values), axis = 1)>0)
marc_s
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Alvaro Fuentes
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5

Total missing:

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

Rows with missing:

sum(map(any, df.isnull()))
ely
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1
# TOTAL number of missing values:
>>> df.isna().sum().sum()
9

# number of ROWS with at least one missing value:
>>> (df.isna().sum(axis=1) > 0).sum()
3

# number of COLUMNS with at least one missing value:
>>> (df.isna().sum(axis=0) > 0).sum()
3

In this example the number of rows and columns with missing values is the same but don't let that confuse you. The point is to use axis=1 or axis=0 in the first sum() method. If you want to see which rows contain any missing records:

>>> df[(df.isna().sum(axis=1) > 0)]

one two three
b   NaN NaN NaN
d   NaN NaN NaN
g   NaN NaN NaN
gojandrooo
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0

sum(df.count(axis=1) < len(df.columns)), the number of rows that have fewer non-nulls than columns.

For example, the following data frame has two rows with missing values.

>>> df = pd.DataFrame({"a":[1, None, 3], "b":[4, 5, None]})
>>> df
    a   b
0   1   4
1 NaN   5
2   3 NaN
>>> df.count(axis=1)
0    2
1    1
2    1
dtype: int64
>>> df.count(axis=1) < len(df.columns)
0    False
1     True
2     True
dtype: bool
>>> sum(df.count(axis=1) < len(df.columns))
2
W.P. McNeill
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0

I think if you just wanna take a look the result, there is a pandas func pandas.DataFrame.count.

So back to this topic, using df.count(axis=1), and u will get the result like this:

a    3
b    0
c    3
d    0
e    3
f    3
g    0
h    3
dtype: int64

It will tell you how many non-NaN parameters in each row. Meanwhile, -(df.count(axis=1) - df.shape[1]) indicates

a    0
b    3
c    0
d    3
e    0
f    0
g    3
h    0
dtype: int64
ruining.z
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0

Regarding counting the number of rows that have somewhere missing values the accepted answer presents

df.shape[0] - df.dropna().shape[0]

But I would rather do the more intuitive (and I also think computationally faster, at least this is what I have read)

len(df) - len(df.dropna())
NeStack
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