根据其他列值对多个数据框列进行分组
Grouping several dataframe columns based on another columns values
我有这个数据框:
refid col2 price1 factor1 price2 factor2 price3 factor3
0 1 a 200 1 180 3 150 10
1 2 b 500 1 450 3 400 10
2 3 c 700 1 620 2 550 5
我需要得到这个输出:
refid col2 price factor
0 1 a 200 1
1 1 b 500 1
2 1 c 700 1
3 2 a 180 3
4 2 b 450 3
5 2 c 620 2
6 3 a 150 10
7 3 b 400 10
8 3 c 550 5
现在我正在尝试使用 df.melt 方法,但无法正常工作,这是代码和当前结果:
df2_melt = df2.melt(id_vars=["refid","col2"],
value_vars=["price1","price2","price3",
"factor1","factor2","factor3"],
var_name="Price",
value_name="factor")
refid col2 price factor
0 1 a price1 200
1 2 b price1 500
2 3 c price1 700
3 1 a price2 180
4 2 b price2 450
5 3 c price2 620
6 1 a price3 150
7 2 b price3 400
8 3 c price3 550
9 1 a factor1 1
10 2 b factor1 1
11 3 c factor1 1
12 1 a factor2 3
13 2 b factor2 3
14 3 c factor2 2
15 1 a factor3 10
16 2 b factor3 10
17 3 c factor3 5
你可以熔化两次然后连接它们:
import pandas as pd
df = pd.DataFrame({'refid': [1, 2, 3], 'col2': ['a', 'b', 'c'],
'price1': [200, 500, 700], 'factor1': [1, 1, 1],
'price2': [180, 450, 620], 'factor2': [3,3,2],
'price3': [150, 400, 550], 'factor3': [10, 10, 5]})
prices = [c for c in df if c.startswith('price')]
factors = [c for c in df if c.startswith('factor')]
df1 = pd.melt(df, id_vars=["refid","col2"], value_vars=prices, value_name='price').drop('variable', axis=1)
df2 = pd.melt(df, id_vars=["refid","col2"], value_vars=factors, value_name='factor').drop('variable', axis=1)
df3 = pd.concat([df1, df2['factor']],axis=1).reset_index().drop('index', axis=1)
print(df3)
这是输出:
refid col2 price factor
0 1 a 200 1
1 2 b 500 1
2 3 c 700 1
3 1 a 180 3
4 2 b 450 3
5 3 c 620 2
6 1 a 150 10
7 2 b 400 10
8 3 c 550 5
由于您有一个带有通用前缀的宽 DataFrame,您可以使用 wide_to_long
:
out = pd.wide_to_long(df, stubnames=['price','factor'],
i=["refid","col2"], j='num').droplevel(-1).reset_index()
输出:
refid col2 price factor
0 1 a 200 1
1 1 a 180 3
2 1 a 150 10
3 2 b 500 1
4 2 b 450 3
5 2 b 400 10
6 3 c 700 1
7 3 c 620 2
8 3 c 550 5
请注意,您的预期输出有一个错误,其中 factor
s 与 refid
s 不一致。
一个选项是pivot_longer from pyjanitor:
# pip install pyjanitor
import janitor
import pandas as pd
(df
.pivot_longer(
index = ['refid', 'col2'],
names_to = '.value',
names_pattern = r"(.+)\d",
sort_by_appearance = True)
)
refid col2 price factor
0 1 a 200 1
1 1 a 180 3
2 1 a 150 10
3 2 b 500 1
4 2 b 450 3
5 2 b 400 10
6 3 c 700 1
7 3 c 620 2
8 3 c 550 5
此特定重塑的想法是,正则表达式中与 .value
配对的任何组都保留为列 header。
我有这个数据框:
refid col2 price1 factor1 price2 factor2 price3 factor3
0 1 a 200 1 180 3 150 10
1 2 b 500 1 450 3 400 10
2 3 c 700 1 620 2 550 5
我需要得到这个输出:
refid col2 price factor
0 1 a 200 1
1 1 b 500 1
2 1 c 700 1
3 2 a 180 3
4 2 b 450 3
5 2 c 620 2
6 3 a 150 10
7 3 b 400 10
8 3 c 550 5
现在我正在尝试使用 df.melt 方法,但无法正常工作,这是代码和当前结果:
df2_melt = df2.melt(id_vars=["refid","col2"],
value_vars=["price1","price2","price3",
"factor1","factor2","factor3"],
var_name="Price",
value_name="factor")
refid col2 price factor
0 1 a price1 200
1 2 b price1 500
2 3 c price1 700
3 1 a price2 180
4 2 b price2 450
5 3 c price2 620
6 1 a price3 150
7 2 b price3 400
8 3 c price3 550
9 1 a factor1 1
10 2 b factor1 1
11 3 c factor1 1
12 1 a factor2 3
13 2 b factor2 3
14 3 c factor2 2
15 1 a factor3 10
16 2 b factor3 10
17 3 c factor3 5
你可以熔化两次然后连接它们:
import pandas as pd
df = pd.DataFrame({'refid': [1, 2, 3], 'col2': ['a', 'b', 'c'],
'price1': [200, 500, 700], 'factor1': [1, 1, 1],
'price2': [180, 450, 620], 'factor2': [3,3,2],
'price3': [150, 400, 550], 'factor3': [10, 10, 5]})
prices = [c for c in df if c.startswith('price')]
factors = [c for c in df if c.startswith('factor')]
df1 = pd.melt(df, id_vars=["refid","col2"], value_vars=prices, value_name='price').drop('variable', axis=1)
df2 = pd.melt(df, id_vars=["refid","col2"], value_vars=factors, value_name='factor').drop('variable', axis=1)
df3 = pd.concat([df1, df2['factor']],axis=1).reset_index().drop('index', axis=1)
print(df3)
这是输出:
refid col2 price factor
0 1 a 200 1
1 2 b 500 1
2 3 c 700 1
3 1 a 180 3
4 2 b 450 3
5 3 c 620 2
6 1 a 150 10
7 2 b 400 10
8 3 c 550 5
由于您有一个带有通用前缀的宽 DataFrame,您可以使用 wide_to_long
:
out = pd.wide_to_long(df, stubnames=['price','factor'],
i=["refid","col2"], j='num').droplevel(-1).reset_index()
输出:
refid col2 price factor
0 1 a 200 1
1 1 a 180 3
2 1 a 150 10
3 2 b 500 1
4 2 b 450 3
5 2 b 400 10
6 3 c 700 1
7 3 c 620 2
8 3 c 550 5
请注意,您的预期输出有一个错误,其中 factor
s 与 refid
s 不一致。
一个选项是pivot_longer from pyjanitor:
# pip install pyjanitor
import janitor
import pandas as pd
(df
.pivot_longer(
index = ['refid', 'col2'],
names_to = '.value',
names_pattern = r"(.+)\d",
sort_by_appearance = True)
)
refid col2 price factor
0 1 a 200 1
1 1 a 180 3
2 1 a 150 10
3 2 b 500 1
4 2 b 450 3
5 2 b 400 10
6 3 c 700 1
7 3 c 620 2
8 3 c 550 5
此特定重塑的想法是,正则表达式中与 .value
配对的任何组都保留为列 header。