您可以用途：

np.random.seed(33454)
stepframe = pd.DataFrame({'a': np.random.randint(1, 200, 20), 
                          'b': np.random.randint(1, 200, 20),
                          'c': np.random.randint(1, 200, 20)})

stepframe[stepframe > 150] *= 10
print (stepframe)

Q1 = stepframe.quantile(0.25)
Q3 = stepframe.quantile(0.75)
IQR = Q3 - Q1

df = stepframe[~((stepframe < (Q1 - 1.5 * IQR)) |(stepframe > (Q3 + 1.5 * IQR))).any(axis=1)]

print (df)
      a    b     c
1   109   50   124
3   137   60  1990
4    19  138   100
5    86   83   143
6    55   23    58
7    78  145    18
8   132   39    65
9    37  146  1970
13   67  148  1880
15  124  102    21
16   93   61    56
17   84   21    25
19   34   52   126

详细信息：

首先通过|创建chain boolean DataFrame：

print (((stepframe < (Q1 - 1.5 * IQR)) | (stepframe > (Q3 + 1.5 * IQR))))
        a      b      c
0   False   True  False
1   False  False  False
2    True  False  False
3   False  False  False
4   False  False  False
5   False  False  False
6   False  False  False
7   False  False  False
8   False  False  False
9   False  False  False
10   True  False  False
11  False   True  False
12  False   True  False
13  False  False  False
14  False   True  False
15  False  False  False
16  False  False  False
17  False  False  False
18  False   True  False
19  False  False  False

然后使用DataFrame.any检查每行至少一个True，最后通过~反转布尔掩码：

print (~((stepframe < (Q1 - 1.5 * IQR)) | (stepframe > (Q3 + 1.5 * IQR))).any(axis=1))
0     False
1      True
2     False
3      True
4      True
5      True
6      True
7      True
8      True
9      True
10    False
11    False
12    False
13     True
14    False
15     True
16     True
17     True
18    False
19     True
dtype: bool

条件已更改的invert解决方案-<至>=和>至<=，按&进行“与”链接，最后按all进行过滤，以检查每行的所有True

print (((stepframe >= (Q1 - 1.5 * IQR)) & (stepframe <= (Q3 + 1.5 * IQR))).all(axis=1))
0     False
1      True
2     False
3      True
4      True
5      True
6      True
7      True
8      True
9      True
10    False
11    False
12    False
13     True
14    False
15     True
16     True
17     True
18    False
19     True
dtype: bool

df = stepframe[((stepframe >= (Q1 - 1.5 * IQR))& (stepframe <= (Q3 + 1.5 * IQR))).all(axis=1)]

您忘记在报价单(['col_name'])中填写列的名称。
正确答案是：

df_out = df_in.loc[(df_in['col_name'] > fence_low) & (df_in['col_name'] < fence_high)]

在使用此函数之前，请确保您的DATAFRAME在“df”变量中，因为我以这种方式构建此函数。如果您有不同的DATAFRAME变量名称，请将“df”替换为您的DATAFRAME变量名称，就是这样。
为什么我构建冗长的函数？？-〉因为我想要每一个信息，如异常值，异常值的索引，这样我就可以看到它是如何与较低和较高的围栏运作。

def outliers(data):
    Q1 = data.quantile(0.25)
    Q3 = data.quantile(0.75)
    IQR = Q3 - Q1

    Lower_fence = Q1 - (1.5*IQR)
    print(f"Lower fence is = {Lower_fence}")

    Higher_fence = Q3 + (1.5*IQR)
    print(f"Higher fence is = {Higher_fence}")
      
    #here i'm taking all Outliers and appending this in Variable "Outlier".
    Outlier =[]
    for i in data:
        if i < Lower_fence:
            Outlier.append(i)
            data.drop(data==i)
        elif i > Higher_fence:
            Outlier.append(i)

    #With the help of "index" function here we are getting all the indexes of Lower_fence and Higher_fence

    Index_Outlier = df[data < Lower_fence ].index  
    Index_Outlier = df[data > Higher_fence].index

    #Here we are converting all the "Outliers" and "Index_Outliers" into string just to see all the data in One line
    #If you do print(Outliers) or print(Outliers_index) you will get every element of data in New Line.

    print(f"\nOutliers = {', '.join([str(x) for x in Outlier])}")
    print(f"\nOutliers_INDEX = {', '.join([str(x) for x in Index_Outlier ])}")

    #here we are seeing before and after shape.

    print(f'\nBEFORE dropping Outlier we have rows = {df.shape[0]}, and columns = {df.shape[1]}')

    df.drop(Index_Outlier,inplace=True)

    print(f'AFTER dropping Outlier we have rows = {df.shape[0]}, and columns = {df.shape[1]}')