Creating rank based on index value in pandas dataframe-CodePudding

I have a data-frame like this

name      qty1  qty2    set_id
0   name0   2   5      1000
1   name1   1   7      1000
2   name2   0   4      1000
3   name3   6   6      1000
4   name4   8   8      1000
5   name5   0   3      1000

I want to update the set_id column like this. (for every nth row(here 3rd row) new set id to be updated.)

name      qty1  qty2    set_id
0   name0   2   5           1
1   name1   1   7           1
2   name2   0   4           2
3   name3   6   6           2
4   name4   8   8           3
5   name5   0   3           3

CodePudding user response：

IIUC, you can use a list comprehension and assign it's return to your set_id column:

df['set_id'] = pd.Series([(i 1) for i in range(len(df)) for n in range(2)])

prints:

        name  qty1  qty2  set_id
0  0   name0     2     5       1
1  1   name1     1     7       1
2  2   name2     0     4       2
3  3   name3     6     6       2
4  4   name4     8     8       3
5  5   name5     0     3       3

You can adjust the number in range(2) to determine your 'nth' row, as you call it.

CodePudding user response：

You can use pd.RangeIndex and repeat:

N = 2  # 3rd row
df['set_id'] = pd.RangeIndex(1, 1 len(df)//N).repeat(N)
print(df)

# Output:
    name  qty1  qty2  set_id
0  name0     2     5       1
1  name1     1     7       1
2  name2     0     4       2
3  name3     6     6       2
4  name4     8     8       3
5  name5     0     3       3

Or the same with np.arange and np.repeat:

N = 2  # 3rd row
df['set_id'] = np.repeat(np.arange(1, 1 len(df)//N), N)
print(df)

# Output:
    name  qty1  qty2  set_id
0  name0     2     5       1
1  name1     1     7       1
2  name2     0     4       2
3  name3     6     6       2
4  name4     8     8       3
5  name5     0     3       3