Question

我有一个pandas.DataFrame有几列，有些列有连续数据，有些列有分类。我一直在尝试按类别分组，然后在每个类别中根据条件（即两个数字之间的值）拆分成数组

这是我写的一个蛮力hackjob做的工作，但我想知道是否有更优雅的方式。

import pandas as pd

df = pd.DataFrame({'Category1' :    [ 0.3,  3.0, 12.4,  7.4,
                                             20.3, 15.0, 10.9, 17.4],
                            'Category2' :    [   0,    0,    1,    0,
                                                 1,    1,    0,    0],
                            'Category3' :    [   1,    2,    3,    4,
                                                 5,    6,    7,    8],
                            'Category4' :  ['foo','bar','fizz','buzz',
                                            'spam','nii','blah','lol'],
                             etc.,                                  })

group_0_5 = df['Category1']<=5.0
group_5_10 = (df['Category1']>5.0) & (df['Category1']<=10.0)
group_10_15 = (df['Category1']>10.0) & (df['Category1']<=15.0)
group_15_20 = (df['Category1']>15.0) & df['Category1']<=20.0)
group_20_25 = (df['Category1']>20.0) & (df['Category1']<=25.0)

state1 = (df['Category2']==1)
state2 = (df['Category2']==0)

count1_state1 = df.loc[group_0_5 & state1]['Category3'].count()
count2_state1 = df.loc[group_5_10 & state1]['Category3'].count()
count3_state1 = df.loc[group_10_15 & state1]['Category3'].count()
count4_state1 = df.loc[group_15_20 & state1]['Category3'].count()
count5_state1 = df.loc[group_20_25 & state1]['Category3'].count()

count1_state2 = df.loc[group_0_5 & state2]['Category3'].count()
count2_state2 = df.loc[group_5_10 & state2]['Category3'].count()
count3_state2 = df.loc[group_10_15 & state2]['Category3'].count()
count4_state2 = df.loc[group_15_20 & state2]['Category3'].count()
count5_state2 = df.loc[group_20_25 & state2]['Category3'].count()

count_array1=[count1_state1, count2_state1, count3_state1, count4_state1, count5_state1]

count_array2=[count1_state2, count2_state2, count3_state2, count4_state2, count5_state2]

print (count_array1)
print (count_array2)

Out [2]:
[nan, nan, 2, 1,   1]
[  2,   1, 1, 1, nan]

Answer 1

我认为您需要cut才能Category2 bins = [-np.inf, 5, 10, 15, 20, 25, np.inf] bins = pd.cut(df['Category1'], bins=bins) mux = pd.MultiIndex.from_product([bins.unique(), df['Category2'].unique()]) a = df.groupby([bins, df['Category2']])['Category3'].count().reindex(mux).unstack(0) print (a) (-inf, 5] (5, 10] (10, 15] (15, 20] (20, 25] 0 2.0 1.0 1.0 1.0 NaN 1 NaN NaN 2.0 NaN 1.0 #select by categories of column Category2 print (a.loc[0].values) [ 2. 1. 1. 1. nan] print (a.loc[1].values) [ nan nan 2. nan 1.]和NaN列进行整理groupby，并在count之前添加缺失值}：

如果需要将fill_value=0替换为reindex，请将参数mux = pd.MultiIndex.from_product([bins.unique(), df['Category2'].unique()]) a = df.groupby([bins, df['Category2']])['Category3'].count() .reindex(mux, fill_value=0) .unstack(0) print (a) (-inf, 5] (5, 10] (10, 15] (15, 20] (20, 25] 0 2 1 1 1 0 1 0 0 2 0 1 print (a.loc[0].values) [2 1 1 1 0] print (a.loc[1].values) [0 0 2 0 1]添加到>>>print(*range(1,11)) 1 2 3 4 5 6 7 8 9 10：

{{1}}

同时检查reindex

Answer 2

使用panda.cut()和pandas.DataFrame.groupby，您可以根据需要收集元素：

<强>代码：

groups = df.groupby(pd.cut(df['Category1'], [0, 5, 10, 15, 20, 25]))

group_size = groups['Category2'].count().values
group_ones = groups['Category2'].sum().values

print(list(group_ones))
print(list(group_size - group_ones))

<强>结果：

[0, 0, 2, 0, 1]
[2, 1, 1, 1, 0]

Answer 3

再次，pd.cut使用groupby和set_index

df = df.groupby([pd.cut(df['Category1'], bins=bins, right = True), 'Category2']).Category3.count().reset_index()
df = df.set_index(['Category1', 'Category2']).unstack().reset_index(-1,drop=True)

count_array1 = df.loc[:, ('Category3', 1)].tolist()
print(count_array1)

[nan, nan, 2.0, nan, 1.0]


count_array2 = df.loc[:, ('Category3', 0)].tolist()
print(count_array2)

[2.0, 1.0, 1.0, 1.0, nan]

按条件对数据帧的行进行排序，并根据其他条件将其拆分为数组

3 个答案: