Question

我有以下数据框df，列为“类”

    Class
0   Individual
1   Group
2   A
3   B
4   C
5   D
6   Group

我想用“其他”替换“组”和“个人”之外的所有内容，所以最终的数据帧是

    Class
0   Individual
1   Group
2   Other
3   Other
4   Other
5   Other
6   Group

数据帧巨大，超过60万行。最佳查找“组”和“个体”以外的值并将其替换为“其他”的最佳方法是什么？

我看到了替换示例，例如：

df['Class'] = df['Class'].replace({'A':'Other', 'B':'Other'})

但是由于我拥有的大量唯一值太多，所以我无法单独执行此操作。我只想使用“组”和“个人”的排除子集。

Answer 1

我认为需要：

df['Class'] = np.where(df['Class'].isin(['Individual','Group']), df['Class'], 'Other')
print (df)
        Class
0  Individual
1       Group
2       Other
3       Other
4       Other
5       Other
6       Group

另一种解决方案（较慢）：

m = (df['Class'] == 'Individual') | (df['Class'] == 'Group')
df['Class'] = np.where(m, df['Class'], 'Other')

另一种解决方案：

df['Class'] = df['Class'].map({'Individual':'Individual', 'Group':'Group'}).fillna('Other')

性能（实际数据取决于替换次数）

#[700000 rows x 1 columns]
df = pd.concat([df] * 100000, ignore_index=True)
#print (df)

In [208]: %timeit df['Class1'] = np.where(df['Class'].isin(['Individual','Group']), df['Class'], 'Other')
25.9 ms ± 485 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [209]: %timeit df['Class2'] = np.where((df['Class'] == 'Individual') | (df['Class'] == 'Group'), df['Class'], 'Other')
120 ms ± 6.63 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [210]: %timeit df['Class3'] = df['Class'].map({'Individual':'Individual', 'Group':'Group'}).fillna('Other')
95.7 ms ± 3.85 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [211]: %timeit df.loc[~df['Class'].isin(['Individual', 'Group']), 'Class'] = 'Other'
97.8 ms ± 6.78 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answer 2

另一种方法可能是：

df.loc[~df['Class'].isin(['Individual', 'Group']), 'Class'] = 'Other'

Answer 3

您可以通过这种方式进行操作

获取唯一商品列表list = df['Class'].unique()
删除您的已知课程list.remove('Individual')。...
然后列出所有其他行df[df.class is in list]
替换类值df[df.class is in list].class = 'Other'

抱歉，此伪伪代码，但原理相同。

Answer 4

您可以使用pd.Series.where：

df['Class'].where(df['Class'].isin(['Individual', 'Group']), 'Other', inplace=True)

print(df)

        Class
0  Individual
1       Group
2       Other
3       Other
4       Other
5       Other
6       Group

与map + fillna相比，这应该是有效的：

df = pd.concat([df] * 100000, ignore_index=True)

%timeit df['Class'].where(df['Class'].isin(['Individual', 'Group']), 'Other')
# 60.3 ms per loop

%timeit df['Class'].map({'Individual':'Individual', 'Group':'Group'}).fillna('Other')
# 133 ms per loop

大熊猫：如果字符串列表中没有该字符串，请用“其他”字符串替换

4 个答案: