Pandas中列的扩展排名

Question

Consider a sample DF:

df = pd.DataFrame(np.random.randint(0,60,size=(10,3)),columns=["a","b","c"])
df["d1"]=["Apple","Mango","Apple","Apple","Mango","Mango","Apple","Mango","Apple","Apple"]
df["d2"]=["Orange","lemon","lemon","Orange","lemon","Orange","lemon","Orange","lemon","Orange"]
df["date"] = ["2002-01-01","2002-01-01","2002-01-01","2002-01-01","2002-02-01","2002-02-01","2002-02-01"]
df["date"] = pd.to_datetime(df["date"])
df

    a   b   c     d1       d2    date
0   7   1   19  Apple   Orange  2002-01-01
1   3   7   17  Mango   lemon   2002-01-01
2   9   6   4   Apple   lemon   2002-01-01
3   0   5   51  Pine    Orange  2002-01-01
4   4   6   8   Apple   lemon   2002-02-01
5   4   3   1   Mango   Orange  2002-02-01
6   2   2   14  Apple   lemon   2002-02-01
7   5   15  10  Mango   Orange  2002-01-01
8   1   2   10  Pine    lemon   2002-02-01
9   2   1   12  Apple   Orange  2002-02-01

Trying to replace column d1 with rank based on Group by column d1 and mean of column c in expanding manner. For example, consider the following first 5 rows:

1. First row by default the value at index 0 , ie Apple will be replaced with 0

2. Second row, index 1 , the value Mango should be replaced by 0 , because considering only the first 2 rows of the DF GROUPED_MEAN for Apple will be 19 and Mango will be 17, so the value Mango at index 1 should be replaced by rank 0 since it has lower grouped mean.

3. Third row, index 2 , the value Apple should be replaced by 0 , because considering only the first 3 rows of the DF GROUPED_MEAN for Apple will be (19+4)/2 and Mango will be 17, so the value Apple at index 2 should be replaced by rank 0 since it has lower grouped mean

4. Fourth row, index 3 , the value Pine should be replaced by 2 , because considering only the first 4 rows of the DF GROUPED_MEAN for Apple will be (19+4)/2 and Mango will be 17, Pine will be 51, since Pine has the highest grouped mean of all the 3 categories- [Apple, Mango, Pine] , Pine will be given rank 2.

5. Fifth row, index 4 , the value Apple should be replaced by 0 , because considering only the first 5 rows of the DF GROUPED_MEAN for Apple will be (19+4+8)/3 and Mango will be 17, Pine will be 51, since Apple has the lowest grouped mean of all the 3 - Apple, Mango, Pine , Apple will be given rank 0.

Expected Value of column d1:

0
0
0
2
0
0
1
0
2
1

Iterative Approach:

def expanding(data,cols):

    copy_df = data.copy(deep=True)
    for i in range(len(copy_df)):
       if i==0:
          copy_df.loc[i,cols]=0
       else:
          op = group_processor(data[:i+1],cols,i)
          copy_df.loc[i,cols]=op
    return copy_df

def group_processor(cut_df,cols,i):

    op=[]
    for each_col in cols:
       temp = cut_df.pivot_table("c",[each_col]).rank(method="dense")-1
       value = cut_df.loc[i,each_col]
       temp = temp.reset_index()
       final_value = temp.loc[temp[each_col]==value,"c"]
       op.append(final_value.values[0])

    return op

expanding(df,["d1"])

I am able do this iteratively through every row of the DF, but the performance is poor for large DFs so any suggestions on a more pandas based approach will be great.

Answer 1

Use Series.expanding with minimum window size of 1 on column c , and use a custom lambda function exp . In this lambda function we use Series.groupby to group the exapnding window w by the column d1 in the original dataframe and transform using mean , finally using Series.rank with method='dense' we calculate the rank:

exp = lambda w: w.groupby(df['d1']).transform('mean').rank(method='dense').iat[-1]
df['d1_new'] = df['c'].expanding(1).apply(exp).sub(1).astype(int)

Result:

# print(df)

   a   b   c     d1      d2        date  d1_new
0  7   1  19  Apple  Orange  2002-01-01       0
1  3   7  17  Mango   lemon  2002-01-01       0
2  9   6   4  Apple   lemon  2002-01-01       0
3  0   5  51   Pine  Orange  2002-01-01       2
4  4   6   8  Apple   lemon  2002-02-01       0
5  4   3   1  Mango  Orange  2002-02-01       0
6  2   2  14  Apple   lemon  2002-02-01       1
7  5  15  10  Mango  Orange  2002-01-01       0
8  1   2  10   Pine   lemon  2002-02-01       2
9  2   1  12  Apple  Orange  2002-02-01       1

Performance:

df.shape
(1000, 7)

%%timeit
exp = lambda w: w.groupby(df['d1']).transform('mean').rank(method='dense').iat[-1]
df['d1_new'] = df['c'].expanding(1).apply(exp).sub(1).astype(int)
3.15 s ± 305 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
expanding(df,["d1"]) # your method
11.9 s ± 449 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)