另一种方法是dropna，检查长度是否相等:

>>> len(df.dropna()) != len(df)
True
>>> 

最好的方法是:

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

原因如下。所以isna()被用来定义isnull()，但这两者当然是相同的。

这甚至比公认的答案还要快，并且涵盖了所有2D熊猫数组。

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

我建议使用值属性作为数组的计算是更快的。

arr = np.random.randn(100, 100)
arr[40, 40] = np.nan
df = pd.DataFrame(arr)

%timeit np.isnan(df.values).any()  # 7.56 µs
%timeit np.isnan(df).any()         # 627 µs
%timeit df.isna().any(axis=None)   # 572 µs

结果:

7.56 µs ± 447 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
627 µs ± 40.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
572 µs ± 15.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

注意:你需要在Jupyter笔记本上运行%timeit才能工作

Df.isnull ().any().any()应该这样做。

下面是另一种有趣的查找null并替换为计算值的方法

    #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