pandas for loop for running average does not work-CodePudding

I tried to make a kind of running average - out of 90 rows, every 3 in column A should make an average that would be the same as those rows in column B. For example: From this:

df = pd.DataFrame( A B 
                   2 0
                   3 0
                   4 0
                   7 0
                   9 0
                   8 0)

to this:

df = pd.DataFrame( A B 
                   2 3
                   3 3
                   4 3
                   7 8
                   9 8
                   8 8)

I tried running this code:

x=0
for i in df['A']:
  if x<90:
    y = (df['A'][x]  df['A'][(x  1)] df['A'][(x  2)])/3
    df['B'][x] = y
    df['B'][(x 1)] = y
    df['B'][(x 2)] = y
    x=x 3
    print(y)

It does print the correct Y But does not change B

I know there is a better way to do it, and if anyone knows - it would be great if they shared it. But the more important thing for me is to understand why what I wrote down doesn't have an effect on the df.

CodePudding user response：

You could group by the index divided by 3, then use transform to compute the mean of those values and assign to B:

df = pd.DataFrame({'A': [2, 3, 4, 7, 9, 8], 'B': [0, 0, 0, 0, 0, 0]})
df['B'] = df.groupby(df.index // 3)['A'].transform('mean')

Output:

Note that this relies on the index being of the form 0,1,2,3,4,.... If that is not the case, you could either reset the index (df.reset_index(drop=True)) or use np.arange(df.shape[0]) instead i.e.

df['B'] = df.groupby(np.arange(df.shape[0]) // 3)['A'].transform('mean')

CodePudding user response：

i = 0
batch_size = 3
df = pd.DataFrame({'A':[2,3,4,7,9,8,9,10],'B':[-1] * 8})
while i < len(df):

  j = min(i batch_size-1,len(df)-1)
  avg =sum(df.loc[i:j,'A'])/ (j-i 1)
  df.loc[i:j,'B'] = [avg] * (j-i 1)
  i =batch_size
df

corner case when len(df) % batch_size != 0 assumes we take the average of the leftover rows.