pandas 将数据帧日期行转换为周末/非周末值
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Convert dataframe date row to a weekend / not weekend value
提问by Raul
I have a dataframe with a DATE row and I need to convert it to a row of value 1 if the date is a weekend day and 0 if it is not.
我有一个带有 DATE 行的数据框，如果日期是周末，我需要将其转换为值为 1 的行，如果不是，则将其转换为 0。
So far, I converted the data to weekdays
到目前为止，我将数据转换为工作日
df['WEEKDAY'] = pandas.to_datetime(df['DATE']).dt.dayofweek
It's there a way to create this "WEEKEND" row without functions?
有没有办法在没有函数的情况下创建这个“WEEKEND”行？
Thanks!
谢谢！
回答by dagrha
Here's the solution I've come up with:
这是我想出的解决方案：
df['WEEKDAY'] = ((pd.DatetimeIndex(df.index).dayofweek) // 5 == 1).astype(float)
Essentially all it does is use integer division (//) to test whether the dayofweekattribute of the DatetimeIndexis less than 5. Normally this would return just a Trueor False, but tacking on the astype(float)at the end returns a 1 or 0 rather than a boolean.
本质上它所做的就是使用整数除法 ( //) 来测试 的dayofweek属性是否DatetimeIndex小于 5。通常这将只返回 a Trueor False，但astype(float)在末尾添加 1 或 0 而不是布尔值。
回答by Vivek Payasi
One more way of getting weekend indicator is by first converting the date column to day of the week and then using those values to get weekend/not weekend. This can be implemented as follows:
获取周末指示器的另一种方法是首先将日期列转换为星期几，然后使用这些值来获取周末/非周末。这可以按如下方式实现：
df['WEEKDAY'] = pandas.to_datetime(df['DATE']).dt.dayofweek # monday = 0, sunday = 6
df['weekend_indi'] = 0 # Initialize the column with default value of 0
df.loc[df['WEEKDAY'].isin([5, 6]), 'weekend_indi'] = 1 # 5 and 6 correspond to Sat and Sun
回答by Daniel Ben-David
One more way of getting weekend indicator is by wherefunction:
获取周末指标的另一种方法是按where功能：
df['WEEKDAY'] = np.where((df['DATE']).dt.dayofweek) < 5,0,1)
