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# Counting the number of missing values (NaNs) in each row of a Pandas DataFrame

schedule Aug 12, 2023
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# Example

Consider the following DataFrame with some `NaN` values:

``` df = pd.DataFrame({"A":[np.nan,3,np.nan],"B":[4,np.nan,5],"C":[np.nan,7,8]}, index=["a","b","c"])df A B Ca NaN 4.0 NaNb 3.0 NaN 7c NaN 5.0 8 ```

## Of each row

To count the number of `NaN` values in each row of `df`:

``` df.isna().sum(axis=1) a 2b 1c 1dtype: int64 ```

### Explanation

Here, the `df.isna()` returns a DataFrame of booleans where `True` indicates entries that are `NaN`:

``` df.isna() A B Ca True False Trueb False True Falsec True False False ```

Internally, `True` is represented by `1` while a `False` is represented by `0`. Therefore, summing up the booleans for each row is equivalent to counting the number of `True` (`NaN` values) per row:

``` df.isna().sum(axis=1) a 2b 1c 1dtype: int64 ```

Here, we must specify `axis=1` so that we are summing each row, and not each column.

## Of a particular row

Consider the same `df` as before:

``` df = pd.DataFrame({"A":[np.nan,3,np.nan],"B":[4,np.nan,5],"C":[np.nan,7,8]}, index=["a","b","c"])df A B Ca NaN 4.0 NaNb 3.0 NaN 7c NaN 5.0 8 ```

To count the number of `NaN` value of just row `a`:

``` df.loc["a"].isna().sum() 2 ```

### Explanation

Here, the DataFrame's `loc` property is first used to extract the row `a`:

``` df.loc["a"] A NaNB 4.0C NaNName: a, dtype: float64 ```

We then use the same tactic as described above to count the number of `NaN`s in this row.

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