一次将列值与另一次比较 pandas 日期时间索引

Question

I have a pandas dataframe with a datetime index and some column, 'value'. I would like to compare the 'value' value at a given time of day to the value at a different time of the same day. Eg compare the 10am value to the 10pm value.

Right now I can get the value at either side using:

mask = df[(df.index.hour == hour)]

the problem is that this returns a dataframe indexed at hour. So doing mask1.value - mask2.value returns Nan's since the indexes are different.

I can get around this in a convoluted way:

out = mask.value.loc["2020-07-15"].reset_index() - mask2.value.loc["2020-07-15"].reset_index() #assuming mask2 is the same as the mask call but at a different hour

but this is tiresome to loop over for a dataset that spans years. (Obviously I could timedelta +=1 in the loop to avoid the hard calls).

I don't actually care if some nan's get into the end result if some, eg 10am, values are missing.

Edit:

Initial dataframe:

index                  values
2020-05-10T10:00:00     23
2020-05-10T11:00:00     20
2020-05-10T12:00:00     5
.....
2020-05-30T22:00:00     8
2020-05-30T23:00:00     8
2020-05-30T24:00:00     9

Expected dataframe:

index        date         newval
  0         2020-05-10     18
.....
  x         2020-05-30     1

where newval is some subtraction of the two different times I described above (eg. the 10am measurement - the 12pm measurement so 23-5 = 18), second entry is made up

it doesn't matter to me if date is a separate column or the index.

Answer 1

A workaround:

mask1 = df[(df.index.hour == hour1)]
mask2 = df[(df.index.hour == hour2)]
out = mask1.values - mask2.values # df.values returns an np array without indices
result_df = pd.DataFrame(index=pd.daterange(start,end), data=out)

It should save you the effort of looping over the dates