python - 只要条件为真，熊猫就会连续累积时间

问题描述

只要“状态” == 1 处于活动状态并且否则“关闭”，就希望有持续时间/时间差异的累积

    timestamp         state
    2020-01-01 00:00:00 0
    2020-01-01 00:00:01 0
    2020-01-01 00:00:02 0
    2020-01-01 00:00:03 1
    2020-01-01 00:00:04 1
    2020-01-01 00:00:05 1
    2020-01-01 00:00:06 1
    2020-01-01 00:00:07 0
    2020-01-01 00:00:08 0
    2020-01-01 00:00:09 0
    2020-01-01 00:00:10 0
    2020-01-01 00:00:11 1
    2020-01-01 00:00:12 1
    2020-01-01 00:00:13 1
    2020-01-01 00:00:14 1
    2020-01-01 00:00:15 1
    2020-01-01 00:00:16 1
    2020-01-01 00:00:17 0
    2020-01-01 00:00:18 0
    2020-01-01 00:00:19 0
    2020-01-01 00:00:20 0

基于类似的问题，我尝试使用 groupby 进行一些操作，但是，当“状态”== 0 时，代码忽略停止执行 timediff。
我还尝试应用 lambda 函数（已注释）但弹出错误说“KeyError：（ '状态'，'发生在索引时间戳'）“知道如何更好地做到这一点吗？

    import numpy as np
    import pandas as pd
    
    dt = pd.date_range('2020-01-01', '2020-01-01 00:00:20',freq='1s')
    s = [0,0,0,1,1,1,1,0,0,0,0,1,1,1,1,1,1,0,0,0,0]
    df = pd.DataFrame({'timestamp': dt,
                       'state': s})
        
    df['timestamp']=pd.to_datetime(df.timestamp, format='%Y-%m-%d %H:%M:%S')
    df['tdiff']=(df.groupby('state').diff().timestamp.values/60)
    #df['tdiff'] = df.apply(lambda x: x['timestamp'].diff().state.values/60 if x['state'] == 1 else 'off')

所需的输出应该是：

    timestamp        state tdiff accum.
    2020-01-01 00:00:00 0   off 0
    2020-01-01 00:00:01 0   off 0
    2020-01-01 00:00:02 0   off 0
    2020-01-01 00:00:03 1   nan 0   
    2020-01-01 00:00:04 1   1.0 1.0
    2020-01-01 00:00:05 1   1.0 2.0
    2020-01-01 00:00:06 1   1.0 3.0
    2020-01-01 00:00:07 0   off 0
    2020-01-01 00:00:08 0   off 0
    2020-01-01 00:00:09 0   off 0
    2020-01-01 00:00:10 0   off 0
    2020-01-01 00:00:11 1   nan 0
    2020-01-01 00:00:12 1   1.0 1.0
    2020-01-01 00:00:13 1   1.0 2.0
    2020-01-01 00:00:14 1   1.0 3.0
    2020-01-01 00:00:15 1   1.0 4.0
    2020-01-01 00:00:16 1   1.0 5.0

标签： pythonpandasperformancenumpytime-series

您可以检查额外的组groupby密钥cumsum

g = df.loc[df['state'].ne(0)].groupby(df['state'].eq(0).cumsum())['timestamp']
s1 = g.diff().dt.total_seconds()
s2 = g.apply(lambda x : x.diff().dt.total_seconds().cumsum())
df['tdiff'] = 'off'
df.loc[df['state'].ne(0),'tdiff'] = s1

df['accum'] = s2
# notice I did not fillna with 0, you can do it with df['accum'].fillna(0,inplace=True)
  
df
Out[53]: 
             timestamp  state tdiff  accum
0  2020-01-01 00:00:00      0   off    NaN
1  2020-01-01 00:00:01      0   off    NaN
2  2020-01-01 00:00:02      0   off    NaN
3  2020-01-01 00:00:03      1   NaN    NaN
4  2020-01-01 00:00:04      1     1    1.0
5  2020-01-01 00:00:05      1     1    2.0
6  2020-01-01 00:00:06      1     1    3.0
7  2020-01-01 00:00:07      0   off    NaN
8  2020-01-01 00:00:08      0   off    NaN
9  2020-01-01 00:00:09      0   off    NaN
10 2020-01-01 00:00:10      0   off    NaN
11 2020-01-01 00:00:11      1   NaN    NaN
12 2020-01-01 00:00:12      1     1    1.0
13 2020-01-01 00:00:13      1     1    2.0
14 2020-01-01 00:00:14      1     1    3.0
15 2020-01-01 00:00:15      1     1    4.0
16 2020-01-01 00:00:16      1     1    5.0
17 2020-01-01 00:00:17      0   off    NaN
18 2020-01-01 00:00:18      0   off    NaN
19 2020-01-01 00:00:19      0   off    NaN
20 2020-01-01 00:00:20      0   off    NaN

python - 只要条件为真，熊猫就会连续累积时间

问题描述

解决方案

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