用python组合列

用python组合列,python,pandas,numpy,dataframe,binning,Python,Pandas,Numpy,Dataframe,Binning,我有一个带有数值的数据框列: df['percentage'].head() 46.5 44.2 100.0 42.12 我希望将列视为bin计数: bins = [0, 1, 5, 10, 25, 50, 100] 如何将结果作为带有值计数的存储箱获取 [0, 1] bin amount [1, 5] etc [5, 10] etc ...... 您可以使用: 或: …然后或合计: 默认情况下,cutreturncategorical Series方法,如Series.valu

我有一个带有数值的数据框列:

df['percentage'].head()
46.5
44.2
100.0
42.12
我希望将列视为bin计数:

bins = [0, 1, 5, 10, 25, 50, 100]
如何将结果作为带有
值计数的存储箱获取

[0, 1] bin amount
[1, 5] etc 
[5, 10] etc 
......
您可以使用:


或:


…然后或合计:


默认情况下,
cut
return
categorical

Series
方法,如
Series.value\u counts()
将使用所有类别,即使数据中不存在某些类别。

使用模块进行加速。 在大数据集(
500k>
)上,
pd.cut
对于装箱数据来说可能非常慢

我在
numba
中使用即时编译编写了自己的函数,这大约比
6x
更快:

from numba import njit

@njit
def cut(arr):
    bins = np.empty(arr.shape[0])
    for idx, x in enumerate(arr):
        if (x >= 0) & (x < 1):
            bins[idx] = 1
        elif (x >= 1) & (x < 5):
            bins[idx] = 2
        elif (x >= 5) & (x < 10):
            bins[idx] = 3
        elif (x >= 10) & (x < 25):
            bins[idx] = 4
        elif (x >= 25) & (x < 50):
            bins[idx] = 5
        elif (x >= 50) & (x < 100):
            bins[idx] = 6
        else:
            bins[idx] = 7
            
    return bins
可选:还可以将其作为字符串映射到存储箱:

a = cut(df['percentage'].to_numpy())

conversion_dict = {1: 'bin1',
                   2: 'bin2',
                   3: 'bin3',
                   4: 'bin4',
                   5: 'bin5',
                   6: 'bin6',
                   7: 'bin7'}

bins = list(map(conversion_dict.get, a))

# ['bin5', 'bin5', 'bin7', 'bin5']

速度比较

# create dataframe of 8 million rows for testing
dfbig = pd.concat([df]*2000000, ignore_index=True)

dfbig.shape

# (8000000, 1)

如果没有
垃圾箱=[0,1,5,10,25,50,100]
,我能说创建5个垃圾箱,它将按平均切割量进行切割吗?例如,我有110条记录,我想把它们分成5个箱子,每个箱子里有22条记录。@qqwww-不确定是否理解,你认为
qcut
@qqwww要做到这一点,pd.cut页面中的示例显示:pd.cut(np.array([1,7,5,4,6,3]),3)将数组分成3等份。@AyanMitra-你认为
df.groupby(pd.cut(df['percentage',bin=bin]).mean()
?谢谢你这个答案帮助了我:)
s = pd.cut(df['percentage'], bins=bins).value_counts()
print (s)
(25, 50]     3
(50, 100]    1
(10, 25]     0
(5, 10]      0
(1, 5]       0
(0, 1]       0
Name: percentage, dtype: int64
s = df.groupby(pd.cut(df['percentage'], bins=bins)).size()
print (s)
percentage
(0, 1]       0
(1, 5]       0
(5, 10]      0
(10, 25]     0
(25, 50]     3
(50, 100]    1
dtype: int64
from numba import njit

@njit
def cut(arr):
    bins = np.empty(arr.shape[0])
    for idx, x in enumerate(arr):
        if (x >= 0) & (x < 1):
            bins[idx] = 1
        elif (x >= 1) & (x < 5):
            bins[idx] = 2
        elif (x >= 5) & (x < 10):
            bins[idx] = 3
        elif (x >= 10) & (x < 25):
            bins[idx] = 4
        elif (x >= 25) & (x < 50):
            bins[idx] = 5
        elif (x >= 50) & (x < 100):
            bins[idx] = 6
        else:
            bins[idx] = 7
            
    return bins
cut(df['percentage'].to_numpy())

# array([5., 5., 7., 5.])
a = cut(df['percentage'].to_numpy())

conversion_dict = {1: 'bin1',
                   2: 'bin2',
                   3: 'bin3',
                   4: 'bin4',
                   5: 'bin5',
                   6: 'bin6',
                   7: 'bin7'}

bins = list(map(conversion_dict.get, a))

# ['bin5', 'bin5', 'bin7', 'bin5']
# create dataframe of 8 million rows for testing
dfbig = pd.concat([df]*2000000, ignore_index=True)

dfbig.shape

# (8000000, 1)
%%timeit
cut(dfbig['percentage'].to_numpy())

# 38 ms ± 616 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%%timeit
bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
pd.cut(dfbig['percentage'], bins=bins, labels=labels)

# 215 ms ± 9.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)