计算熊猫数据框的列中某个字符的出现

Question

I have a dataframe with the following structure

Debtor_ID    | Loan_ID    | Pattern_of_payments
Uncle Sam      Loan1        11111AAA11555
Uncle Sam      Loan2        11222A339999
Uncle Joe      Loan3        1111111111111
Uncle Joe      Loan4        111222222233333
Aunt Annie     Loan5        1
Aunt Chloe     Loan6        555555555

Each character in the column "Pattern_of_payments" marks on-time payment (like 1, for instance) or delay(like all the rest). What I want to do is count the number of occurrence of each character in each row of "Pattern_of_payments" column and assign that number to a respective column in dataframe like this:

Debtor_ID    | Loan_ID    | On_time_payment    | 1_29_days_delay    | 30_59_days_delay    | 60_89_days_delay    | 90_119_days_delay    | Over_120_days_delay    | Bailiff_prosecution
Uncle Sam      Loan1        7                    3                    0                     0                     0                      3                        0
Uncle Sam      Loan2        2                    1                    3                     2                     0                      3                        4
Uncle Joe      Loan3        13                   0                    0                     0                     0                      0                        0
Uncle Joe      Loan4        3                    0                    7                     4                     0                      0                        0
Aunt Annie     Loan5        1                    0                    0                     0                     0                      0                        0
Aunt Chloe     Loan6        0                    0                    0                     0                     0                      9                        0

My code accomplishes the task in this manner:

list_of_counts_of_1 = []
list_of_counts_of_A = []
list_of_counts_of_2 = []
list_of_counts_of_3 = []
list_of_counts_of_4 = []
list_of_counts_of_5 = []
list_of_counts_of_8 = []
list_of_counts_of_9 = []
for value in df_account.Pattern_of_payments.values:
    iter_string = str(value)
    count1 = iter_string.count("1")
    countA = iter_string.count("A")
    count2 = iter_string.count("2")
    count3 = iter_string.count("3")
    count4 = iter_string.count("4")
    count5 = iter_string.count("5")
    count8 = iter_string.count("8")
    count9 =  iter_string.count("9")
    list_of_counts_of_1.append(count1)
    list_of_counts_of_A.append(countA)
    list_of_counts_of_2.append(count2)
    list_of_counts_of_3.append(count3)
    list_of_counts_of_4.append(count4)
    list_of_counts_of_5.append(count5)
    list_of_counts_of_9.append(count9)
df_account["On_time_payment"] = list_of_counts_of_1
df_account["1_29_days_delay"] = list_of_counts_of_A
df_account["30_59_days_delay"] = list_of_counts_of_2
df_account["60_89_days_delay"] = list_of_counts_of_3
df_account["90_119_days_delay"] = list_of_counts_of_4
df_account["Over_120_days_delay"] = list_of_counts_of_5
df_account["Bailiff_prosecution"] = list_of_counts_of_9

I realize that my code isn't "pythonic" at all. There has to be a way to express this in a way more succinct manner (maybe even some fancy one-liner). Please advise how would the best practice for coding look like?

Answer 1

First step is create DataFrame by Counter in list comprehension, then use reindex for add missing categories and change order of columns, rename columns by dict and add to original DataFrame by join :

from collections import Counter

df1 = pd.DataFrame([Counter(list(x)) for x in df['Pattern_of_payments']], index=df.index)
order = list('1A23459')

d = {'1': "On_time_payment",
     'A': "1_29_days_delay",
     '2':"30_59_days_delay",
     '3':"60_89_days_delay",
     '4':"90_119_days_delay",
     '5':"Over_120_days_delay",
     '9':"Bailiff_prosecution"}

df2 = df1.fillna(0).astype(int).reindex(columns=order, fill_value=0).rename(columns=d)
df = df.join(df2)

---

print (df)
    Debtor_ID Loan_ID Pattern_of_payments  On_time_payment  1_29_days_delay  \
0   Uncle Sam   Loan1       11111AAA11555                7                3   
1   Uncle Sam   Loan2        11222A339999                2                1   
2   Uncle Joe   Loan3       1111111111111               13                0   
3   Uncle Joe   Loan4     111222222233333                3                0   
4  Aunt Annie   Loan5                   1                1                0   
5  Aunt Chloe   Loan6           555555555                0                0   

   30_59_days_delay  60_89_days_delay  90_119_days_delay  Over_120_days_delay  \
0                 0                 0                  0                    3   
1                 3                 2                  0                    0   
2                 0                 0                  0                    0   
3                 7                 5                  0                    0   
4                 0                 0                  0                    0   
5                 0                 0                  0                    9   

   Bailiff_prosecution  
0                    0  
1                    4  
2                    0  
3                    0  
4                    0  
5                    0