从 R 到 python pandas：按顺序创建 id 键系列

Question

I want to create an id key for a series on string that repeats in one column. The first ten rows should be id #1, the next ten #2 and so on. In R, this is simple and I get the expected result with dplyr.

R:

library(tidyverse)

question = c('q1', 'q2', 'q3', 'q4', 'q5', 'q1', 'q2', 'q3', 'q4', 'q5', 'q1', 'q2', 'q3', 'q4', 'q5')
answer <- c('a1', 'a2', 'a3', 'a4', 'a5', 'a1', 'a2', 'a3', 'a4', 'a5', 'a1', 'a2', 'a3', 'a4', 'a5')

df <- data_frame(question, answer)

# A tibble: 15 x 2
   question answer
   <chr>    <chr> 
 1 q1       a1    
 2 q2       a2    
 3 q3       a3    
 4 q4       a4    
 5 q5       a5    
 6 q1       a1    
 7 q2       a2    
 8 q3       a3    
 9 q4       a4    
10 q5       a5    
11 q1       a1    
12 q2       a2    
13 q3       a3    
14 q4       a4    
15 q5       a5 

If we run just a group_by and a mutate to add a key to the series we get what I want:

df2 <- df %>% 
  group_by(question) %>% 
  mutate(id = row_number())

# A tibble: 15 x 3
# Groups:   question [5]
   question answer    id
   <chr>    <chr>  <int>
 1 q1       a1         1
 2 q2       a2         1
 3 q3       a3         1
 4 q4       a4         1
 5 q5       a5         1
 6 q1       a1         2
 7 q2       a2         2
 8 q3       a3         2
 9 q4       a4         2
10 q5       a5         2

And I finish with:

df2 <- df %>% 
  group_by(question) %>% 
  mutate(id = row_number()) %>% 
  spread(question, answer) 

# final table:
# A tibble: 3 x 6
      id    q1    q2    q3    q4    q5   
      <int> <chr> <chr> <chr> <chr> <chr>
    1     1 a1    a2    a3    a4    a5   
    2     2 a1    a2    a3    a4    a5   
    3     3 a1    a2    a3    a4    a5 

Python:

Now, I can't figure out how to get the same result in Pandas. I have tried groupby and merge but no luck.

import pandas as pd

data = {'question': ['question one', 'question two', 
                 'question three', 'question four', 
                 'question five', 'question one', 
                 'question two', 'question three', 
                 'question four', 'question five', 
                 'question one', 'question two', 
                 'question three', 'question four', 'question five'], 
    'answer':['answer one', 'answer two', 'answer three', 
              'answer four', 'answer five', 'answer one', 
              'answer two', 'answer three', 'answer four', 
              'answer five', 'answer one', 'answer two', 
              'answer three', 'answer four', 'answer five']}

df = pd.DataFrame(data)

Using merge and rest_index() it reorders the rows and assigns an id on a new order and that is not what I want:

df2 = df.merge(df.drop_duplicates('question').reset_index(), on='question')

          question      answer_x  index      answer_y
0     question one    answer one      0    answer one
1     question one    answer one      0    answer one
2     question one    answer one      0    answer one
3     question two    answer two      1    answer two
4     question two    answer two      1    answer two
5     question two    answer two      1    answer two

Using groupby I get a mess that is also not what I want:

df['id'] = df.groupby('question').ngroup()

          question        answer  id
0     question one    answer one   2
1     question two    answer two   4
2   question three  answer three   3
3    question four   answer four   1
4    question five   answer five   0
5     question one    answer one   2
6     question two    answer two   4
7   question three  answer three   3
8    question four   answer four   1
9    question five   answer five   0

How do I get the same output as with dplyr? Edit: To add more details, I need the output to be like dplyr is giving me as this is part of an automated system.

Answer 1

ngroup is the number of the group, not the number within a group. As the docs explain, the complement of this is given by cumcount .

Roughly, you can use assign for mutate , groupby/cumcount for row_number , and pivot for your spread :

In [306]: df.assign(id=df.groupby("question").cumcount()).pivot("id", "question", "answer")
Out[306]: 
question  q1  q2  q3  q4  q5
id                          
0         a1  a2  a3  a4  a5
1         a1  a2  a3  a4  a5

and toss in a reset_index() if you want id to be a column.

Unfortunately, I guess to really match the expected output, we'd have to guarantee the order. There are several open tickets on github about how the automatic sorting is inconvenient, but we can do it manually. We'll switch back to the English text:

In [327]: d2 = df.assign(id=df.groupby("question").cumcount()).pivot("id", "question", "answer")

In [328]: d2.reindex(df.question.drop_duplicates(), axis=1)
Out[328]: 
question question one question two question three question four question five
id                                                                           
0          answer one   answer two   answer three   answer four   answer five
1          answer one   answer two   answer three   answer four   answer five

Answer 2

With datar , you can replicate it easily as you did in R:

>>> from datar.all import c, f, tibble, group_by, mutate, row_number, pivot_wider
>>> 
>>> question = c('q1', 'q2', 'q3', 'q4', 'q5', 'q1', 'q2', 'q3', 'q4', 'q5', 'q1', 'q2', 'q3', 'q4', 'q
5')
>>> answer = c('a1', 'a2', 'a3', 'a4', 'a5', 'a1', 'a2', 'a3', 'a4', 'a5', 'a1', 'a2', 'a3', 'a4', 'a5'
)
>>> 
>>> df = tibble(question, answer)
>>> df
   question answer
0        q1     a1
1        q2     a2
2        q3     a3
3        q4     a4
4        q5     a5
5        q1     a1
6        q2     a2
7        q3     a3
8        q4     a4
9        q5     a5
10       q1     a1
11       q2     a2
12       q3     a3
13       q4     a4
14       q5     a5

>>> df2 = (df >>
...   group_by(f.question) >>
...   mutate(id = row_number()))
>>> 
>>> df2
   question answer  id
0        q1     a1   1
1        q2     a2   1
2        q3     a3   1
3        q4     a4   1
4        q5     a5   1
5        q1     a1   2
6        q2     a2   2
7        q3     a3   2
8        q4     a4   2
9        q5     a5   2
10       q1     a1   3
11       q2     a2   3
12       q3     a3   3
13       q4     a4   3
14       q5     a5   3
[Groups: ['question'] (n=5)]

>>> df2 = (df >>
...   group_by(f.question) >>
...   mutate(id = row_number()) >>
...   pivot_wider(names_from=f.question, values_from=f.answer))
>>> 
>>> df2
   id  q1  q2  q3  q4  q5
0   1  a1  a2  a3  a4  a5
1   2  a1  a2  a3  a4  a5
2   3  a1  a2  a3  a4  a5

I am the author of the package. Feel free to submit issues if you have any questions.

Answer 3

I know the question is about how to obtain a solution in python, still, I will leave this solution using data.table and reshape2 .

library(data.table)
library(reshape2)
setDT(df)[,new := (1:.N), by = question]
dcast(df, new ~ question, value.var = "answer")

   new q1 q2 q3 q4 q5
1:   1 a1 a2 a3 a4 a5
2:   2 a1 a2 a3 a4 a5
3:   3 a1 a2 a3 a4 a5