使用 Pandas 对列进行分组
- 2024-12-02 08:41:00
- admin 原创
- 139
问题描述:
我有一个包含数值的数据框列:
df['percentage'].head()
46.5
44.2
100.0
42.12
我想看到bin 计数的列:
bins = [0, 1, 5, 10, 25, 50, 100]
我怎样才能获得带有其值计数的箱子结果?
[0, 1] bin amount
[1, 5] etc
[5, 10] etc
...
解决方案 1:
您可以使用pandas.cut
:
bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = pd.cut(df['percentage'], bins)
print (df)
percentage binned
0 46.50 (25, 50]
1 44.20 (25, 50]
2 100.00 (50, 100]
3 42.12 (25, 50]
bins = [0, 1, 5, 10, 25, 50, 100]
labels = [1,2,3,4,5,6]
df['binned'] = pd.cut(df['percentage'], bins=bins, labels=labels)
print (df)
percentage binned
0 46.50 5
1 44.20 5
2 100.00 6
3 42.12 5
或者numpy.searchsorted
:
bins = [0, 1, 5, 10, 25, 50, 100]
df['binned'] = np.searchsorted(bins, df['percentage'].values)
print (df)
percentage binned
0 46.50 5
1 44.20 5
2 100.00 6
3 42.12 5
...然后value_counts
或groupby
和聚合size
:
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
默认cut
返回categorical
。
Series
诸如此类的方法Series.value_counts()
将使用所有类别,即使数据中不存在某些类别,也要进行分类操作。
解决方案 2:
使用Numba模块来加速。
对于大型数据集(超过 500k),pd.cut
数据分箱的速度可能非常慢。
我使用 Numba 中的即时编译编写了自己的函数,速度大约提高了六倍:
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.])
可选:您还可以将其作为字符串映射到箱子:
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 a 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 ± standard deviation 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 ± standard deviation of 7 runs, 10 loops each)
解决方案 3:
使用 Numpy 的便捷选项
np.digitize是一个方便快捷的选择:
import pandas as pd
import numpy as np
df = pd.DataFrame({'x': [1,2,3,4,5]})
df['y'] = np.digitize(df['x'], bins=[3,5]) # convert column to bin
print(df)
返回
x y
0 1 0
1 2 0
2 3 1
3 4 1
4 5 2
解决方案 4:
我们还可以使用np.select
:
bins = [0, 1, 5, 10, 25, 50, 100]
df['groups'] = (np.select([df['percentage'].between(i, j, inclusive='right')
for i,j in zip(bins, bins[1:])],
[1, 2, 3, 4, 5, 6]))
输出:
percentage groups
0 46.50 5
1 44.20 5
2 100.00 6
3 42.12 5
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