如何根据 boolean 条件更改 pandas dataframe 中的单元格

Question

我必须关注 dataframe . 我成功地删除了这个 dataframe 闰年的所有 2 月 29 天，因为我打算按“一年中的一天”列（使用.dt.dayofyear 创建）分组，我决定忽略闰年的额外一天。 现在，为了按“一年中的一天”列进行分组，如果这一天是三月的第一天或更晚，我必须从闰年的天数中减去 1。 否则，闰年将有 366 天而不是 355 天（即使在删除闰日之后）。

这是我的代码：

clim_rec = pd.read_csv("daily_climate_records.csv")
clim_rec['Date'] = pd.to_datetime(clim_rec['Date']) # converting "Date" column from string into datetime format

# Let's drop all leaping days by masking all Feb 29 days
feb_29_mask = ~((clim_rec.Date.dt.month == 2) & (clim_rec.Date.dt.day == 29))
clim_rec = clim_rec.where(feb_29_mask).dropna()

# Let's add new column with the "day of year" in order to group by this column
clim_rec['Day of year'] = clim_rec['Date'].dt.dayofyear
print(clim_rec.head())
#print('---------------------------------------------------')
# Now, if the year is a leap year and the dayofyear is greater than the dayofyear of Feb-29
# we subtract 1 from dayofyear. After doing that we will get values 1-365 for dayofyear
leap_year_mask = (clim_rec.Date.dt.year % 4 == 0) & ((clim_rec.Date.dt.year % 100 != 0)
                 |(clim_rec.Date.dt.year % 400 == 0)) & (clim_rec.Date.dt.month >=3)

clim_rec['Day of year'] = clim_rec['Day of year'].apply(lambda x: x-1) # this line is not correct

我的问题是：如何修改附加代码的最后一行，以便仅对根据 boolean 掩码条件为真的特定行应用减法

Answer 1

通过掩码将DataFrame.loc用于 select 行，更好/更快地减去1而不是apply避免循环（因为在引擎盖下应用循环）：

clim_rec.loc[leap_year_mask, 'Day of year'] -= 1 

像这样工作：

clim_rec.loc[leap_year_mask, 'Day of year'] = clim_rec.loc[leap_year_mask, 'Day of year']-1

Answer 2

这对你有用吗？ 氪。

clim_rec['mask'] = leaf_year_mask
clim_rec['Day of year'] =  clim_rec.apply(lambda x: x['Day of year']-1 if x['mask'] else x['Day of year'])