我有一个重新采样的 df:
import pandas as pd
import numpy as np
nat = np.datetime64('NaT')
df = pd.DataFrame({"Time": [nat, nat, nat, '2020-04-09 06:45:38.559871', '2020-04-09 06:45:38.559871', nat, nat, nat, '2020-04-09 06:50:16.268515', '2020-04-09 06:50:16.268515'],
"Power": [0, 0, 0, 4200, 4200, 0, 0, 0, 4200, 4200],
"Total Energy": [5300, 5300, 5300, 5500, 5600, 5600, 5600, 5600, 5900, 6100],
"ID": ['-', '-', '-', 1, 1, '-', '-', '-', 2, 2],
"Energy": [0, 0, 0, 200, 300, 0, 0, 0, 300, 500]},
index=pd.date_range(start = "2020-04-09 6:45", periods = 10, freq = 'T'))
df['Time'] = pd.to_datetime(df['Time'])
df['Power'] = pd.to_numeric(df['Power'], errors = 'ignore')
df['Total Energy'] = pd.to_numeric(df['Total Energy'], errors = 'coerce')
df['ID'] = pd.to_numeric(df['ID'], errors = 'coerce')
df['Energy'] = pd.to_numeric(df['Energy'], errors = 'coerce')
df
输出:
Time Power Total Energy ID Energy
2020-04-09 06:45:00 NaT 0 5300 NaN 0
2020-04-09 06:46:00 NaT 0 5300 NaN 0
2020-04-09 06:47:00 NaT 0 5300 NaN 0
2020-04-09 06:48:00 2020-04-09 06:45:38.559871 4200 5500 1.0 200
2020-04-09 06:49:00 2020-04-09 06:45:38.559871 4200 5600 1.0 300
2020-04-09 06:50:00 NaT 0 5600 NaN 0
2020-04-09 06:51:00 NaT 0 5600 NaN 0
2020-04-09 06:52:00 NaT 0 5600 NaN 0
2020-04-09 06:53:00 2020-04-09 06:50:16.268515 4200 5900 2.0 300
2020-04-09 06:54:00 2020-04-09 06:50:16.268515 4200 6100 2.0 500
我必须填写 df.index < df['Time']
所在的行(四舍五入)和df['Time'] == NaT
如下:
df.index == df['Time']: df['Power'] = 0, df['Total Energy']
保持不变,df['ID'] = df['ID'] of the df['Time'], df['Energy'] = 0
df['Power'] = df['Energy']
@ df['Time'] / ((df['Time'] (rounded) - df.index)/60)
, df['Energy'] = df['Power'] * 1/60, df['Total Energy'] = df['Total Energy'].shift(1) + df['Energy'], df['ID'] = df['ID']
的 df['Time']
这里是期望的结果:
Time Power Total Energy ID Energy
2020-04-09 06:45:00 NaT 0 5300 NaN 0
2020-04-09 06:46:00 2020-04-09 06:45:38.559871 0 5300 1.0 0
2020-04-09 06:47:00 2020-04-09 06:45:38.559871 6000 5400 1.0 100
2020-04-09 06:48:00 2020-04-09 06:45:38.559871 4200 5500 1.0 200
2020-04-09 06:49:00 2020-04-09 06:45:38.559871 4200 5600 1.0 300
2020-04-09 06:50:00 NaT 0 5600 NaN 0
2020-04-09 06:51:00 2020-04-09 06:50:16.268515 0 5600 2.0 0
2020-04-09 06:52:00 2020-04-09 06:50:16.268515 9000 5750 2.0 150
2020-04-09 06:53:00 2020-04-09 06:50:16.268515 4200 5900 2.0 300
2020-04-09 06:54:00 2020-04-09 06:50:16.268515 4200 6100 2.0 500
专栏df['Time']
也可以更改为舍入值:
Time Power Total Energy ID Energy
2020-04-09 06:45:00 NaT 0 5300 NaN 0
2020-04-09 06:46:00 2020-04-09 06:46:00 0 5300 1.0 0
2020-04-09 06:47:00 2020-04-09 06:46:00 6000 5400 1.0 100
2020-04-09 06:48:00 2020-04-09 06:46:00 4200 5500 1.0 200
2020-04-09 06:49:00 2020-04-09 06:46:00 4200 5600 1.0 300
2020-04-09 06:50:00 NaT 0 5600 NaN 0
2020-04-09 06:51:00 2020-04-09 06:51:00 0 5600 2.0 0
2020-04-09 06:52:00 2020-04-09 06:51:00 9000 5750 2.0 150
2020-04-09 06:53:00 2020-04-09 06:51:00 4200 5900 2.0 300
2020-04-09 06:54:00 2020-04-09 06:51:00 4200 6100 2.0 500
感谢您的帮助:)
编辑
舍入 df['Time']
我发现了这个:
df['Time'] = df['Time'].dt.ceil('1min')
编辑2
调整列df['Time']
我的处理如下:
dates = df['Time'].unique()
for date in dates:
for index, row in df.iterrows():
if index == date:
df.loc[index, 'Time'] = date
我如何获得 df[ID]
相应列的 ( df['Time']
) 我还不知道。
我还按如下方式填充了行:
#scheme for filling the nan-values
s = df['Time'].ffill()
x = df['Time'].bfill()
g = df['Time'].mask(s.eq(x), s)
#Filling time
df['Time'] = df['Time'].groupby(g).ffill()
#Filling ID
df3['ID'] = df2['ID'].groupby(df2['Time']).bfill()
输出:
Time Power Total Energy ID Energy
2020-04-09 06:45:00 NaT 0 5300 NaN 0
2020-04-09 06:46:00 2020-04-09 06:46:00 0 5300 1.0 0
2020-04-09 06:47:00 2020-04-09 06:46:00 0 5300 1.0 0
2020-04-09 06:48:00 2020-04-09 06:46:00 4200 5500 1.0 200
2020-04-09 06:49:00 2020-04-09 06:46:00 4200 5600 1.0 300
2020-04-09 06:50:00 NaT 0 5600 NaN 0
2020-04-09 06:51:00 2020-04-09 06:51:00 0 5600 2.0 0
2020-04-09 06:52:00 2020-04-09 06:51:00 0 5600 2.0 0
2020-04-09 06:53:00 2020-04-09 06:51:00 4200 5900 2.0 300
2020-04-09 06:54:00 2020-04-09 06:51:00 4200 6100 2.0 500
仍然缺少:df['Power']/ df['Energy']
的值和 df['Total Energy']
必须如上所述进行计算和更改。
最佳答案
鉴于您想要的输出以及您在评论中告诉我的内容,我这样做了:
time_bfill = df['Time'].bfill()
df['Time2'] = df['Time'].mask(df.index.to_series().ge(time_bfill), time_bfill)
id_bfill = df['ID'].bfill()
df['ID2'] = df['ID'].mask(df.index.to_series().ge(time_bfill), id_bfill)
df['Energy2'] = df['Energy'].mask((df['Time'].isnull()) & (df['Time2'].notna()) & (df.index.to_series().ne(df['Time2'])), np.NaN)
df['Energy3'] = df['Energy2'].interpolate(limit_direction='both', limit_area='inside')
df['Power2'] = df['Power'].mask(df['Power'] == 0, 60 * df['Energy3'])
df['Total Energy2'] = df['Total Energy'].mask(df['Power'] == 0, df['Total Energy'] + df['Energy3'])
df
得到这个数据框:
考虑具有最大后缀的列。我将它们留在这里供您查看中间步骤。
您可以调整该代码以避免一些中间列,但要小心,因为在某些情况下其他列需要原始值,在它们之后生成。
摆脱那些中间列的快速方法是在最后运行它:
df[['Time', 'Power', 'Total Energy', 'ID', 'Energy']] = df[['Time2', 'Power2', 'Total Energy2', 'ID2', 'Energy3']]
df.drop(['Time2', 'Power2', 'Total Energy2', 'ID2', 'Energy3', 'Energy2'], axis=1, inplace=True)
df
你得到:
https://stackoverflow.com/questions/63778229/