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将数据帧[duplicate]中的值从字符串映射到int

mapping pandas python

Aaraeus · 技术社区 · 6 年前

我有一个叫做 df_base 看起来像这样。如你所见,有一个专栏叫做 Sex male 或 female . 我想把这些值分别映射到0和1。

+---+-------------+----------+--------+---------------------------------------------------+--------+-----+-------+-------+------------------+---------+-------+----------+
|   | PassengerId | Survived | Pclass |                       Name                        |  Sex   | Age | SibSp | Parch |      Ticket      |  Fare   | Cabin | Embarked |
+---+-------------+----------+--------+---------------------------------------------------+--------+-----+-------+-------+------------------+---------+-------+----------+
| 0 |           1 |        0 |      3 | Braund, Mr. Owen Harris                           | male   |  22 |     1 |     0 | A/5 21171        |    7.25 | NaN   | S        |
| 1 |           2 |        1 |      1 | Cumings, Mrs. John Bradley (Florence Briggs Th... | female |  38 |     1 |     0 | PC 17599         | 71.2833 | C85   | C        |
| 2 |           3 |        1 |      3 | Heikkinen, Miss. Laina                            | female |  26 |     0 |     0 | STON/O2. 3101282 |   7.925 | NaN   | S        |
| 3 |           4 |        1 |      1 | Futrelle, Mrs. Jacques Heath (Lily May Peel)      | female |  35 |     1 |     0 | 113803           |    53.1 | C123  | S        |
| 4 |           5 |        0 |      3 | Allen, Mr. William Henry                          | male   |  35 |     0 |     0 | 373450           |    8.05 | NaN   | S        |
+---+-------------+----------+--------+---------------------------------------------------+--------+-----+-------+-------+------------------+---------+-------+----------+

我在StackOverflow上看到了一些方法,但我想知道执行以下映射最有效的方法是什么:

+---------+---------+
| Old Sex | New Sex |
+---------+---------+
| male    |       0 |
| female  |       1 |
| female  |       1 |
| female  |       1 |
| male    |       0 |
+---------+---------+

我用这个:

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

... 但我不由得觉得这有点伪劣。有没有更好的办法?还有使用 .loc 但是它在数据帧的行之间循环,所以效率较低,对吧?

3 回复 | 直到 6 年前

jezrael 6 年前

我认为这是更好/更快的使用方法 map 只要查字典就行了 male 和 female Sex :

df_base['Sex'] = df_base['Sex'].map(dict(zip(['male','female'],[0,1]))

df_base['Sex'] = df_base['Sex'].map({'male': 0,'female': 1})

女性的 和 男性的 值被布尔掩码转换为整数 True/False 到 1,0 :

df_base['Sex'] = (df_base['Sex'] == 'female').astype(int)

性能 :

np.random.seed(2019)

import perfplot    

def ma(df):
    df = df.copy()
    df['Sex_new'] = df['Sex'].map({'male': 0,'female': 1})
    return df

def rep1(df):
    df = df.copy()
    df['Sex'] = df['Sex'].replace(['male','female'],[0,1])
    return df

def nwhere(df):
    df = df.copy()
    df['Sex_new'] = np.where(df['Sex'] == 'male', 0, 1)
    return df

def mask1(df):
    df = df.copy()
    df['Sex_new'] = (df['Sex'] == 'female').astype(int)
    return df

def mask2(df):
    df = df.copy()
    df['Sex_new'] = (df['Sex'].values == 'female').astype(int)
    return df


def make_df(n):
    df = pd.DataFrame({'Sex': np.random.choice(['male','female'], size=n)})

    return df

perfplot.show(
    setup=make_df,
    kernels=[ma,  rep1, nwhere, mask1, mask2],
    n_range=[2**k for k in range(2, 18)],
    logx=True,
    logy=True,
    equality_check=False,  # rows may appear in different order
    xlabel='len(df)')

如果仅替换2个值,则速度最慢 replace , numpy.where, map and mask .values .
也都取决于数据,所以最好用真实数据测试。

Niels Henkens 6 年前

我的直觉会建议 .map() ,但我将您的解决方案与map进行了比较,基于一个包含1500个随机男性/女性值的数据帧。

%timeit df_base['Sex_new'] = df_base['Sex'].map({'male': 0,'female': 1})
1000 loops, best of 3: 653 Âµs per loop

%timeit df_base['Sex_new'] = df_base['Sex'].replace(['male','female'],[0,1])
1000 loops, best of 3: 968 Âµs per loop

.map() ...!

~~所以基于这个例子,你的“劣质”解决方案似乎比~~

编辑

pygo的解决方案:

%timeit df_base['Sex_new'] = np.where(df_base['Sex'] == 'male', 0, 1)
1000 loops, best of 3: 331 Âµs per loop

.astype(int) :

%timeit df_base['Sex_new'] = (df_base['Sex'] == 'female').astype(int)
1000 loops, best of 3: 388 Âµs per loop

.map() 和 .replace() .

Karn Kumar 6 年前

另一个解决方案,你可以使用 np.where

仅举一个数据帧示例:

>>> df
      Sex
0    male
1  female
2  female
3  female
4    male

根据条件创建新列 new_Sex

>>> df['new_Sex'] = np.where(df['Sex'] == 'male', 0, 1)
>>> df
      Sex  new_Sex
0    male        0
1  female        1
2  female        1
3  female        1
4    male        0

>>> df['new_Sex'] = np.where(df['Sex'] != 'male', 1, 0)
>>> df
      Sex  new_Sex
0    male        0
1  female        1
2  female        1
3  female        1
4    male        0