我正在嘗試為資料集中的每個日期提取第二個最接近的日期。下面是一些示例資料
asOfDate maturityDate value
0 2022-09-01 2022-10-01 57.273
1 2022-09-01 2022-11-01 55.861
2 2022-09-01 2022-12-01 59.231
3 2022-09-01 2023-01-01 59.305
4 2022-09-01 2023-02-01 58.081
5 2022-09-01 2023-03-01 51.198
6 2022-09-01 2023-04-01 44.532
7 2022-09-01 2023-05-01 38.955
8 2022-09-01 2023-06-01 37.901
9 2022-09-01 2023-07-01 40.461
10 2022-09-01 2023-08-01 40.126
11 2022-09-01 2023-09-01 40.221
12 2022-09-01 2023-10-01 40.455
13 2022-09-01 2023-11-01 41.652
14 2022-09-01 2023-12-01 42.450
15 2022-09-01 2024-01-01 45.057
16 2022-09-01 2024-02-01 42.459
17 2022-09-01 2024-03-01 37.544
18 2022-09-01 2024-04-01 29.901
19 2022-09-01 2024-05-01 25.300
20 2022-09-01 2024-06-01 22.360
21 2022-09-01 2024-07-01 22.544
22 2022-09-01 2024-08-01 23.931
我有一個 asOfDates 串列，其中有幾個到期日期。我目前正在使用此代碼來提取每個 asOfDate 的最小（或滾動前一個月）的成熟日期和值。
df_final[df_final['maturityDate']==df_final.groupby(['asOfDate'])['maturityDate'].transform(min)]
以上內容非常適合我的需求，但現在我想，而不是獲得最小的日期，而是接收每個日期的第二小日期。我試圖得到這個第二小的，但我收到一個錯誤，指出“錯誤：只能比較標簽相同的系列物件”。下面是我嘗試過的輸出我想要的日期的方法，但是當我嘗試重新索引資料框時，我得到了錯誤。
df_final.groupby(['asOfDate'])['maturityDate'].nsmallest(2).groupby(['asOfDate']).last().reset_index()
df_final[df_final['maturityDate']==df_final.groupby(['asOfDate'])['maturityDate'].nsmallest(2).groupby(['asOfDate']).last().reset_index()]
上面的第一個輸出如下，這是期望的結果以及這里的值有什么想法嗎？
asOfDate maturityDate
0 2022-09-01 2022-11-01
1 2022-09-02 2022-11-01
2 2022-09-05 2022-11-01
3 2022-09-06 2022-11-01
4 2022-09-07 2022-11-01
5 2022-09-08 2022-11-01
6 2022-09-09 2022-11-01
7 2022-09-12 2022-11-01
8 2022-09-13 2022-11-01
9 2022-09-14 2022-11-01
10 2022-09-15 2022-11-01
11 2022-09-16 2022-12-01
12 2022-09-19 2022-12-01
13 2022-09-20 2022-12-01
14 2022-09-21 2022-12-01
15 2022-09-22 2022-12-01
16 2022-09-23 2022-12-01
17 2022-09-26 2022-12-01
18 2022-09-27 2022-12-01
19 2022-09-28 2022-12-01
20 2022-09-29 2022-12-01
21 2022-09-30 2022-12-01
22 2022-10-03 2022-12-01
23 2022-10-04 2022-12-01
24 2022-10-05 2022-12-01
25 2022-10-06 2022-12-01
26 2022-10-07 2022-12-01
27 2022-10-10 2022-12-01
28 2022-10-11 2022-12-01
29 2022-10-12 2022-12-01
uj5u.com熱心網友回復：
import pandas as pd
columns = ["asOfDate", "maturityDate", "value"]
data = [
["2022-09-01", "2022-10-01", 57.273]
,["2022-09-01", "2022-11-01", 55.861]
,["2022-09-01", "2022-12-01", 59.231]
,["2022-09-01", "2023-01-01", 59.305]
,["2022-09-01", "2023-02-01", 58.081]
,["2022-09-01", "2023-03-01", 51.198]
,["2022-09-01", "2023-04-01", 44.532]
,["2022-09-01", "2023-05-01", 38.955]
,["2022-09-01", "2023-06-01", 37.901]
,["2022-09-01", "2023-07-01", 40.461]
,["2022-09-01", "2023-08-01", 40.126]
,["2022-09-01", "2023-09-01", 40.221]
,["2022-09-01", "2023-10-01", 40.455]
,["2022-09-01", "2023-11-01", 41.652]
,["2022-09-01", "2023-12-01", 42.450]
,["2022-09-01", "2024-01-01", 45.057]
,["2022-09-01", "2024-02-01", 42.459]
,["2022-09-01", "2024-03-01", 37.544]
,["2022-09-01", "2024-04-01", 29.901]
,["2022-09-01", "2024-05-01", 25.300]
,["2022-09-01", "2024-06-01", 22.360]
,["2022-09-01", "2024-07-01", 22.544]
,["2022-09-01", "2024-08-01", 23.931]
]
df = pd.DataFrame.from_records(data=data, columns=columns)
for col in ["asOfDate", "maturityDate"]:
df[col] = pd.to_datetime(df[col])
df.sort_values(["asOfDate","maturityDate"]).groupby("asOfDate").agg({"maturityDate":lambda x: x.shift(-1).values[0]})
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標籤：Python熊猫数据框