如何将Elif与Pandas DataFrame一起使用

Question

我有一个如下所示的销售数据框

    Company         Sales
    MC                360.0
    MC                340.0
    MC                338.5
    MC                335.5
    MC                235.0
    MC                235.0
    MC                234.0
    MC                127.0
    MC                121.0
    MC                120.5

我想基于Sales列的值创建一个新列，并且我的输出表应该像

    CompanyCode     ActivityDate    Category
    MC                360.0         Fast Mover
    MC                340.0         Fast Mover
    MC                338.5         Fast Mover
    MC                335.5         Fast Mover
    MC                235.0         Medium Fast Mover
    MC                235.0         Medium Fast Mover
    MC                234.0         Medium Fast Mover
    MC                127.0         Slow Mover
    MC                121.0         Slow Mover
    MC                120.5         Slow Mover

我尝试了elif声明

if df['Sales']>=300:
   df['Category'] = 'Fast Movers'
elif (df['Sales']>=200) & (df['Sales'] < 300) :
   df['Category'] = 'Medium Fast Movers'
else:
   df['Category'] = 'Slow Movers'

不知道我的方法是否正确。 感谢您的帮助和努力。

Answer 1

将lambda函数（或命名函数）与apply一起apply

df.Category = df.Sales.apply(
    lambda x: 'Fast Movers' if x >=300 else ('Medium Fast Movers' if x >= 200 else 'Slow Movers')
)

代替lambda，可以这样定义一个命名函数：

def move_speed(x):
    if x >= 300:
        return 'Fast Movers'
    if x >= 200:
        return 'Medium Fast Movers'
    return 'Slow Movers'

或者，您可以使用loc分别分配每个条件

df.loc[df.Sales > 300, 'Category'] = 'Fast Movers'
df.loc[df.Sales < 200, 'Category'] = 'Slow Movers'
df.Category.fillna('Medium Fast Movers', inplace=True)

两者都会产生这个数据帧：

  Company  Sales            Category
0      MC  360.0         Fast Movers
1      MC  340.0         Fast Movers
2      MC  338.5         Fast Movers
3      MC  335.5         Fast Movers
4      MC  235.0  Medium Fast Movers
5      MC  235.0  Medium Fast Movers
6      MC  234.0  Medium Fast Movers
7      MC  127.0         Slow Movers
8      MC  121.0         Slow Movers
9      MC  120.5         Slow Movers

Answer 2

您可以使用np.select()

df["Category"] = np.select([df.Sales>=300, df.Sales<200], 
                           ["Fast Movers", "Slow Movers"], 
                           default="Medium Fast Movers")

Answer 3

这是使用第二个数据帧和pd.IntervalIndex的解决方案。 如果您有很多范围，这可以使代码更整洁。

设定

ranges = pd.DataFrame({
    'START': [0, 200, 300],
    'STOP': [200, 300, float('inf')],
    'CLASS': ['Slow Workers', 'Medium Workers', 'Fast Workers']
})

#                          CLASS  START        STOP
# [0.0, 200.0)      Slow Workers      0  200.000000
# [200.0, 300.0)  Medium Workers    200  300.000000
# [300.0, inf)      Fast Workers    300         inf

ranges.index = pd.IntervalIndex.from_arrays(ranges.START, ranges.STOP, closed='left')

map

df['Category'] = df.Sales.map(ranges.CLASS)

# Result

  Company  Sales        Category
0      MC  360.0    Fast Workers
1      MC  340.0    Fast Workers
2      MC  338.5    Fast Workers
3      MC  335.5    Fast Workers
4      MC  235.0  Medium Workers
5      MC  235.0  Medium Workers
6      MC  234.0  Medium Workers
7      MC  127.0    Slow Workers
8      MC  121.0    Slow Workers
9      MC  120.5    Slow Workers