您可以使用groupby傳遞日期時間年。但首先我們需要刪除(過濾掉)不符合標準的數據。還要確保你的日期是日期時間。
此代碼將檢查該月是否等於12月(12),並且該日是大於或等於25(即每年的最後7天)。如果你想要一年的最後一週,你可以看看Wen's lambda函數。
data = '''\
2017-12-25 52.99 13018070.0 52.370 53.0600 51.4000
2017-01-04 52.86 12556860.0 50.770 53.3400 50.7300
2017-01-03 50.29 15794400.0 48.800 50.3000 48.4700
2016-12-30 46.75 13593420.0 48.365 48.4000 46.3600
2016-12-29 47.77 11728250.0 48.440 48.8600 47.1800
2016-12-28 48.51 14636340.0 50.580 50.7300 48.4700
2016-12-27 50.43 5594876.0 49.690 50.5500 49.6500
2016-12-23 49.59 6966559.0 49.250 49.7200 48.9900
2016-12-22 49.44 10918300.0 50.320 50.5500 49.1711
2016-12-21 50.34 9279635.0 49.820 50.4400 49.6700
2016-12-20 49.53 9533020.0 48.990 49.7900 48.9100
2016-12-19 48.55 10323930.0 47.450 48.6700 47.4300'''
import io
import pandas as pd
df = pd.read_csv(io.StringIO(data), sep='\s+', header=None, parse_dates=[0])
df = df[df[0].dt.month.eq(12) & df[0].dt.day.le(25)] # remove data
# Groupby year according to: https://stackoverflow.com/a/11397052/7386332
for idx, dfx in df.groupby(df[0].map(lambda x: x.year)):
print('Dataframe containing {}\'s last week:'.format(idx))
print(dfx)
print()
打印
Dataframe containing 2016's last week:
0 1 2 3 4 5
7 2016-12-23 49.59 6966559.0 49.25 49.72 48.9900
8 2016-12-22 49.44 10918300.0 50.32 50.55 49.1711
9 2016-12-21 50.34 9279635.0 49.82 50.44 49.6700
10 2016-12-20 49.53 9533020.0 48.99 49.79 48.9100
11 2016-12-19 48.55 10323930.0 47.45 48.67 47.4300
Dataframe containing 2017's last week:
0 1 2 3 4 5
0 2017-12-25 52.99 13018070.0 52.37 53.06 51.4
的數據其實我想過這樣的事情,但是這將取決於歷年,對不對?這就是爲什麼我認爲> = 25會起作用。取決於OP真正想要的。 –
@AntonvBR如何定義最後一週在這裏:-) – Wen