Aks*_*kar 2 python time-series pandas
我有一个时间序列数据如下:
print(df)
ric datel timel val
0 xyz 2017-01-01 09:00:00 2
1 xyz 2017-01-01 09:04:00 5
2 xyz 2017-01-01 09:37:00 6
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现在我必须将缺失的时间戳填充到09:45:00.
预期输出:
ric datel timel val
0 xyz 2017-01-01 09:00:00 2
1 xyz 2017-01-01 09:01:00 nan
2 xyz 2017-01-01 09:02:00 nan
3 xyz 2017-01-01 09:03:00 nan
4 xyz 2017-01-01 09:04:00 5
...
...
37 xyz 2017-01-01 09:37:00 6
...
...
45 xyz 2017-01-01 09:45:00 nan
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我试过的:
df1=df.resample("1 min", on ='datel').first()
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它给出的输出为:
ric datel timel val
datel
2017-01-01 xyz 2017-01-01 09:00:00 2
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也尝试过,pd.date_range但它主要适用于日期时间列。我有两个不同的日期和时间列。有没有办法在不将日期和列合并到日期时间的情况下实现这一目标?
主要思想是reindex由times 创建的使用date_range:
df['timel'] = pd.to_datetime(df['timel']).dt.time
start = pd.to_datetime(str(df['timel'].min()))
end = pd.to_datetime('09:45:00')
dates = pd.date_range(start=start, end=end, freq='1Min').time
#print (dates)
df = df.set_index('timel').reindex(dates).reset_index().reindex(columns=df.columns)
cols = df.columns.difference(['val'])
df[cols] = df[cols].ffill()
print (df.head())
ric datel timel val
0 xyz 2017-01-01 09:00:00 2.0
1 xyz 2017-01-01 09:01:00 NaN
2 xyz 2017-01-01 09:02:00 NaN
3 xyz 2017-01-01 09:03:00 NaN
4 xyz 2017-01-01 09:04:00 5.0
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类似的解决方案resample:
df['timel'] = pd.to_datetime(df['timel'])
#if missing row with 09:45:00 add it
if not (df['timel'] == pd.to_datetime('09:45:00')).any():
df.loc[len(df.index), 'timel'] = pd.to_datetime('09:45:00')
df=df.set_index('timel').resample("1min").first().reset_index().reindex(columns=df.columns)
cols = df.columns.difference(['val'])
df[cols] = df[cols].ffill()
df['timel'] = df['timel'].dt.time
print (df.head())
ric datel timel val
0 xyz 2017-01-01 09:00:00 2.0
1 xyz 2017-01-01 09:01:00 NaN
2 xyz 2017-01-01 09:02:00 NaN
3 xyz 2017-01-01 09:03:00 NaN
4 xyz 2017-01-01 09:04:00 5.0
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