groupby cumsum(或cumcount)与周期性数据

huangapple go评论67阅读模式
英文:

groupby cumsum (or cumcount) with cyclical data

问题

ID SWITCH Cum. Count
A ON 1
A ON 2
A ON 3
A OFF 1
A OFF 2
A OFF 3
A ON 1
A ON 2
A ON 3
... ...
B ON 1
B ON 2
B OFF 1
B OFF 2
B OFF 3
B ON 1
B ON 2
B ON 3
英文:

I have a dataframe that looks like

ID SWITCH
A ON
A ON
A ON
A OFF
A OFF
A OFF
A ON
A ON
A ON
... ...
B ON
B ON
B ON
B OFF
B OFF
B OFF
B ON
B ON
B ON

Column 'SWITCH' is cyclical data and I'd like to count the number of ON and OFF for each cycle like this:

ID SWITCH Cum. Count
A ON 1
A ON 2
A ON 3
A OFF 1
A OFF 2
A OFF 3
A ON 1
A ON 2
A ON 3
... ...
B ON 1
B ON 2
B OFF 1
B OFF 2
B OFF 3
B ON 1
B ON 2
B ON 3

I'd tried cumsum or cumcount but it didn't reset the count when the next 'ON' cycle has come (it keeps counting on the number from the previous cycle).

What can I do?

答案1

得分: 1

尝试将差异的累加和也添加进去:

switch_blocsk = df['SWITCH'].ne(df['SWITCH'].shift()).cumsum()

df['cum.count'] = df.groupby(['ID', switch_blocks]).cumcount().add(1)
英文:

Try put in the cumsum of the difference as well:

switch_blocsk = df['SWITCH'].ne(df['SWITCH'].shift()).cumsum()

df['cum.count'] = df.groupby(['ID', switch_blocks]).cumcount().add(1)

答案2

得分: 1

你需要创建一个新列,表示'SWITCH'列的变化,然后可以使用'groupby'来执行累积计数。

结果:

ID SWITCH SWITCH_CHANGE Cum. Count
0 A ON 1
1 A ON 0
2 A ON 0
3 A OFF 1
4 A OFF 0
5 A OFF 0
6 A ON 1
7 A ON 0
8 A ON 0
9 B ON 0
10 B ON 0
11 B ON 0
12 B OFF 1
13 B OFF 0
14 B OFF 0
15 B OFF 0
16 B OFF 0
17 B OFF 0
英文:

You need to create a new column which indicates the change in the 'SWITCH' column, then you can use 'groupby' to perform the cumulative count.

import pandas as pd

# Create sample data
df = pd.DataFrame({'ID': ['A'] * 9 + ['B'] * 9,
                   'SWITCH': ['ON'] * 3 + ['OFF'] * 3 + ['ON'] * 3 + ['ON'] * 3 + ['OFF'] * 3 + ['OFF'] * 3})

df['SWITCH_CHANGE'] = (df['SWITCH'] != df['SWITCH'].shift()).astype(int)

df['Cum. Count'] = df.groupby(['ID', df.SWITCH_CHANGE.cumsum()])['SWITCH'].cumcount() + 1

print(df)

Result:

ID SWITCH SWITCH_CHANGE Cum. Count
0 A ON 1 1
1 A ON 0 2
2 A ON 0 3
3 A OFF 1 1
4 A OFF 0 2
5 A OFF 0 3
6 A ON 1 1
7 A ON 0 2
8 A ON 0 3
9 B ON 0 1
10 B ON 0 2
11 B ON 0 3
12 B OFF 1 1
13 B OFF 0 2
14 B OFF 0 3
15 B OFF 0 4
16 B OFF 0 5
17 B OFF 0 6

huangapple
  • 本文由 发表于 2023年2月6日 12:36:33
  • 转载请务必保留本文链接:https://go.coder-hub.com/75357378.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定