计算 pandas dataframe 中 ID 的出现次数

Question

I have aa few dataframes, a few thousand rows each that look similar to this:

heifers_df

       id   y     ins               
200316157 123 2004121 
200316157 456 2004121 
200316157 789 2004121 
200519776 456 2007234 
200519776 789 2007234 
200812334 123 2010333 
200812334 789 2010333 
200812334 345 2010333 
200812334 567 2010333 

I want to use python (pandas or numphy? ) to count the occurrence if each ID, both total occurrences (T) and the number of each occurrence (No) :

heifers_df

       id    y      ins  T  No          
200316157  123  2004121  3   1
200316157  456  2004121  3   2
200316157  789  2004121  3   3
200519776  456  2007234  2   1
200519776  789  2007234  2   2
200812334  123  2010333  4   1
200812334  789  2010333  4   2
200812334  345  2010333  4   3
200812334  567  2010333  4   4

I've gotten help with this problem in Fortran Counting frequency of variables in text data in Fortran But now I'm trying to accomplish the same in python.

Based on the Fortran code and my beginner knowledge of python and pandas this is what I've tried doing with the first dataframe:

i1 = 0
# set i0, i1
#  i0: line where specific user id starts
#  i1: line where specific user id ends
for i in range(len(heifers_df)) :
    i0 = i1 + 1
    same_id = True
    while same_id == True :
        heifers_df.loc[
            heifers_df["id"[i]] != heifers_df["id"[i0]],     #How do I reference each row within the column?
            same_id ] = False
    i1 = i
    heifers_df["T"] = i1-i0+1
    heifers_df["No"] = i-i0+1

But when I run this I get an error:

....  heifers_df["id"[i]] != heifers_df["id"[i0]],
     KeyError: 'i'

Am I going in the wrong direction with this?

I've tried to search for similar problems and I've seen group by and counting operations but I haven't seen one that glues the result to the IDs in questions and counts each one. Any help would be much appreciated.

Answer 1

IIUC, if all unique id's can be sorted into contiguous blocks.

df['T'] = df.groupby('id')['id'].transform('count')
df['No'] = df.groupby('id')['id'].cumcount() + 1
df

Output:

          id    y      ins  T  No
0  200316157  123  2004121  3   1
1  200316157  456  2004121  3   2
2  200316157  789  2004121  3   3
3  200519776  456  2007234  2   1
4  200519776  789  2007234  2   2
5  200812334  123  2010333  4   1
6  200812334  789  2010333  4   2
7  200812334  345  2010333  4   3
8  200812334  567  2010333  4   4