使用系列。使用Series.fillna映射

如果您的列包含的字符串多于female和male，则Series。map在这种情况下将失败，因为它将为其他值返回NaN。

这就是为什么我们要用fillna来连接它

.map失败的例子:

df = pd.DataFrame({'female':['male', 'female', 'female', 'male', 'other', 'other']})

   female
0    male
1  female
2  female
3    male
4   other
5   other

df['female'].map({'female': '1', 'male': '0'})

0      0
1      1
2      1
3      0
4    NaN
5    NaN
Name: female, dtype: object

对于正确的方法，我们使用fillna进行链映射，所以我们用原始列的值填充NaN:

df['female'].map({'female': '1', 'male': '0'}).fillna(df['female'])

0        0
1        1
2        1
3        0
4    other
5    other
Name: female, dtype: object

你也可以使用apply和。get i.e.。

"菲玛"等于"菲玛"克里奥耳语(——“画”:0,可说是家常便饭”:1,.get):

w = pd.DataFrame({'female':['female','male','female']})
print(w)

Dataframe w:

   female
0  female
1    male
2  female

使用apply替换字典中的值:

w['female'] = w['female'].apply({'male':0, 'female':1}.get)
print(w)

结果:

   female
0       1
1       0
2       1 

注意:如果数据帧中所有可能的列的值都是在字典else中定义的，那么apply with dictionary应该被使用，对于那些没有在字典中定义的列，它将为空。

轻微的变化:

w.female.replace(['male', 'female'], [1, 0], inplace=True)

为了更一般地回答这个问题，使它适用于更多的用例，而不仅仅是OP要求的用例，可以考虑这个解决方案。我使用jfs的解决方案来帮助我。在这里，我们创建了两个相互帮助的函数，无论您是否知道确切的替换都可以使用它们。

import numpy as np
import pandas as pd


class Utility:

    @staticmethod
    def rename_values_in_column(column: pd.Series, name_changes: dict = None) -> pd.Series:
        """
        Renames the distinct names in a column. If no dictionary is provided for the exact name changes, it will default
        to <column_name>_count. Ex. female_1, female_2, etc.

        :param column: The column in your dataframe you would like to alter.
        :param name_changes: A dictionary of the old values to the new values you would like to change.
        Ex. {1234: "User A"} This would change all occurrences of 1234 to the string "User A" and leave the other values as they were.
        By default, this is an empty dictionary.
        :return: The same column with the replaced values
        """
        name_changes = name_changes if name_changes else {}
        new_column = column.replace(to_replace=name_changes)
        return new_column

    @staticmethod
    def create_unique_values_for_column(column: pd.Series, except_values: list = None) -> dict:
        """
        Creates a dictionary where the key is the existing column item and the value is the new item to replace it.
        The returned dictionary can then be passed the pandas rename function to rename all the distinct values in a
        column.
        Ex. column ["statement"]["I", "am", "old"] would return
        {"I": "statement_1", "am": "statement_2", "old": "statement_3"}

        If you would like a value to remain the same, enter the values you would like to stay in the except_values.
        Ex. except_values = ["I", "am"]
        column ["statement"]["I", "am", "old"] would return
        {"old", "statement_3"}

        :param column: A pandas Series for the column with the values to replace.
        :param except_values: A list of values you do not want to have changed.
        :return: A dictionary that maps the old values their respective new values.
        """
        except_values = except_values if except_values else []
        column_name = column.name
        distinct_values = np.unique(column)
        name_mappings = {}
        count = 1
        for value in distinct_values:
            if value not in except_values:
                name_mappings[value] = f"{column_name}_{count}"
                count += 1
        return name_mappings

对于OP的用例，使用它非常简单

w["female"] = Utility.rename_values_in_column(w["female"], name_changes = {"female": 0, "male":1}

然而，要知道您想要重命名的数据帧中所有不同的惟一值并不总是那么容易。在我的例子中，列的字符串值是散列值，因此它们损害了可读性。我所做的是使用create_unique_values_for_column函数将这些散列值替换为更可读的字符串。

df["user"] = Utility.rename_values_in_column(
    df["user"],
    Utility.create_unique_values_for_column(df["user"])
)

这将改变我的用户列值从["1a2b3c"， "a12b3c"，"1a2b3c"]到["user_1"， "user_2"， "user_1]。比较起来容易多了，对吧?

pandas中还有一个叫做factorize的函数，您可以使用它自动完成这类工作。它将标签转换为数字:['male'， 'female'， 'male'] ->[0,1,0]。更多信息请看这个答案。

使用系列。使用Series.fillna映射

如果您的列包含的字符串多于female和male，则Series。map在这种情况下将失败，因为它将为其他值返回NaN。

这就是为什么我们要用fillna来连接它

.map失败的例子:

df = pd.DataFrame({'female':['male', 'female', 'female', 'male', 'other', 'other']})

   female
0    male
1  female
2  female
3    male
4   other
5   other

df['female'].map({'female': '1', 'male': '0'})

0      0
1      1
2      1
3      0
4    NaN
5    NaN
Name: female, dtype: object

对于正确的方法，我们使用fillna进行链映射，所以我们用原始列的值填充NaN:

df['female'].map({'female': '1', 'male': '0'}).fillna(df['female'])

0        0
1        1
2        1
3        0
4    other
5    other
Name: female, dtype: object