英文:
How to swap month and day values in a DataFrame?
问题
我有一个DataFrame，其中第一列包含日期和时间信息。但是对我来说，这些信息被错误地识别了。当我从数据库中提取信息并将数据发送到DataFrame表时，它看起来像是我“交换了位置” - “天”和“月”。“天”和“月”被颠倒了 - 天的值必须与月的值相同。“月”和“天”被颠倒了 - 月的值必须是天的值。如何交换并使天像月一样，月像天一样？在DataFrame表中，一列被识别为类型 - datetime。但是月和日的识别是错误的 - 它们被交换了。如何在DataFrame中交换月和日的值？
--
之前
id datetime temp_narvoz q_virob0 1 2023-01-02 10:00:00 1.4 0.6883311 2 2023-01-02 11:00:00 1.4 0.8008672 3 2023-01-02 12:00:00 1.4 0.8107463 4 2023-01-02 13:00:00 1.4 0.8055224 5 2023-01-02 14:00:00 2.1 0.8029791 Jan. 2, 2023, 10 a.m. 1.4 0.68833133730446482 Jan. 2, 2023, 11 a.m. 1.4 0.80086745757795523 Jan. 2, 2023, noon 1.4 0.81074620698228564 Jan. 2, 2023, 1 p.m. 1.4 0.80552227302393035 Jan. 2, 2023, 2 p.m. 2.1 0.80297860551124016 Jan. 2, 2023, 3 p.m. 2.1 0.78540978390137767 Jan. 2, 2023, 4 p.m. 2.1 0.79503601496943958 Jan. 2, 2023, 5 p.m. 1.6 0.82963867616285089 Jan. 2, 2023, 6 p.m. 1.6 0.8300509596498510 Jan. 2, 2023, 7 p.m. 1.6 0.856999553506682111 Jan. 2, 2023, 8 p.m. -0.5 0.871051538796228512 Jan. 2, 2023, 9 p.m. -0.5 0.86416424945612813 Jan. 2, 2023, 10 p.m. -0.5 0.851456768154977814 Jan. 2, 2023, 11 p.m. 0.3 0.807884754791256715 Feb. 2, 2023, midnight 0.3 0.7834063591629548
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之后
id datetime temp_narvoz q_virob0 1 2023-02-01 10:00:00 1.4 0.6883311 2 2023-02-01 11:00:00 1.4 0.8008672 3 2023-02-01 12:00:00 1.4 0.8107463 4 2023-02-01 13:00:00 1.4 0.8055224 5 2023-02-01 14:00:00 2.1 0.8029791 Feb. 1, 2023, 10 a.m. 1.4 0.68833133730446482 Feb. 1, 2023, 11 a.m. 1.4 0.80086745757795523 Feb. 1, 2023, noon 1.4 0.81074620698228564 Feb. 1, 2023, 1 p.m. 1.4 0.80552227302393035 Feb. 1, 2023, 2 p.m. 2.1 0.80297860551124016 Feb. 1, 2023, 3 p.m. 2.1 0.78540978390137767 Feb. 1, 2023, 4 p.m. 2.1 0.79503601496943958 Feb. 1, 2023, 5 p.m. 1.6 0.82963867616285089 Feb. 1, 2023, 6 p.m. 1.6 0.8300509596498510 Feb. 1, 2023, 7 p.m. 1.6 0.856999553506682111 Feb. 1, 2023, 8 p.m. -0.5 0.871051538796228512 Feb. 1, 2023, 9 p.m. -0.5 0.86416424945612813 Feb. 1, 2023, 10 p.m. -0.5 0.851456768154977814 Feb. 1, 2023, 11 p.m. 0.3 0.807884754791256715 Feb. 1, 2023, midnight 0.3 0.7834063591629548
英文:
I have a DataFrame in which the first column contains information in the form of a date and time.
But for me this information is not recognized correctly.
When I took information from the database and sent the data to the DataFrame table, it looks like I "swapped places" - "day" and "month".
"Day" and "month" are reversed - the day value must be the same value as the month.
"Month" and "day" are reversed - the month value must be a day value.
How can I flip and make the day be like a month, and the month like a day?
In a DataFrame table, a column is recognized as a type - datetime.
But the recognition of the month and day is wrong - they are swapped.
How to swap month and day values in a DataFrame?
now the date time is being recognized sodf['datetime'] = pd.to_datetime(df["datetime"].dt.strftime('%Y-%m-%d'))and the real location of the month and day isdf['datetime'] = pd.to_datetime(df["datetime"].dt.strftime('%Y-%d-%m'))
--
before
id datetime temp_narvoz q_virob0 1 2023-01-02 10:00:00 1.4 0.6883311 2 2023-01-02 11:00:00 1.4 0.8008672 3 2023-01-02 12:00:00 1.4 0.8107463 4 2023-01-02 13:00:00 1.4 0.8055224 5 2023-01-02 14:00:00 2.1 0.8029791 Jan. 2, 2023, 10 a.m. 1.4 0.68833133730446482 Jan. 2, 2023, 11 a.m. 1.4 0.80086745757795523 Jan. 2, 2023, noon 1.4 0.81074620698228564 Jan. 2, 2023, 1 p.m. 1.4 0.80552227302393035 Jan. 2, 2023, 2 p.m. 2.1 0.80297860551124016 Jan. 2, 2023, 3 p.m. 2.1 0.78540978390137767 Jan. 2, 2023, 4 p.m. 2.1 0.79503601496943958 Jan. 2, 2023, 5 p.m. 1.6 0.82963867616285089 Jan. 2, 2023, 6 p.m. 1.6 0.8300509596498510 Jan. 2, 2023, 7 p.m. 1.6 0.856999553506682111 Jan. 2, 2023, 8 p.m. -0.5 0.871051538796228512 Jan. 2, 2023, 9 p.m. -0.5 0.86416424945612813 Jan. 2, 2023, 10 p.m. -0.5 0.851456768154977814 Jan. 2, 2023, 11 p.m. 0.3 0.807884754791256715 Feb. 2, 2023, midnight 0.3 0.7834063591629548
--
after
id datetime temp_narvoz q_virob0 1 2023-02-01 10:00:00 1.4 0.6883311 2 2023-02-01 11:00:00 1.4 0.8008672 3 2023-02-01 12:00:00 1.4 0.8107463 4 2023-02-01 13:00:00 1.4 0.8055224 5 2023-02-01 14:00:00 2.1 0.8029791 Feb. 1, 2023, 10 a.m. 1.4 0.68833133730446482 Feb. 1, 2023, 11 a.m. 1.4 0.80086745757795523 Feb. 1, 2023, noon 1.4 0.81074620698228564 Feb. 1, 2023, 1 p.m. 1.4 0.80552227302393035 Feb. 1, 2023, 2 p.m. 2.1 0.80297860551124016 Feb. 1, 2023, 3 p.m. 2.1 0.78540978390137767 Feb. 1, 2023, 4 p.m. 2.1 0.79503601496943958 Feb. 1, 2023, 5 p.m. 1.6 0.82963867616285089 Feb. 1, 2023, 6 p.m. 1.6 0.8300509596498510 Feb. 1, 2023, 7 p.m. 1.6 0.856999553506682111 Feb. 1, 2023, 8 p.m. -0.5 0.871051538796228512 Feb. 1, 2023, 9 p.m. -0.5 0.86416424945612813 Feb. 1, 2023, 10 p.m. -0.5 0.851456768154977814 Feb. 1, 2023, 11 p.m. 0.3 0.807884754791256715 Feb. 1, 2023, midnight 0.3 0.7834063591629548
答案1
得分: 0
import pandas as pd
df = pd.DataFrame({'date': ['11/02/2022', '12/02/2023', '13/02/2023']})
df['date'] = pd.to_datetime(df['date'])
df['date'] = df['date'].apply(lambda x: x.replace(day=x.month,
month=x.day))
print(df)
英文:
import pandas as pddf = pd.DataFrame({'date': ['11/02/2022', '12/02/2023', '13/02/2023']})df['date'] = pd.to_datetime(df['date'])df['date'] = df['date'].apply(lambda x: x.replace(day=x.month,month=x.day))print(df)
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