根据R中的分类变量比较两组

Question

I have created df which contains more than 8,000 firm years

gvkey = company id

fam = dummy (equals 1 if firm is family firm)

industry = categorial variable

   gvkey   fam  industry
1   1004    0     6
2   1004    0     6
3   1004    0     6
4   1004    0     6
5   1004    0     6
6   1013    0     4
7   1013    0     4
8   1013    0     4
9   1013    0     4
10  1013    0     4
11  1013    0     4
12  1045    0     5
13  1045    0     5
14  1045    0     5
15  1045    0     5
16  1045    0     5
17  1045    0     5
18  1072    0     4
19  1072    0     4
20  1072    0     4
21  1072    0     4
22  1072    0     4
23  1076    1     9
24  1076    1     9
25  1076    1     9
26  1076    1     9
27  1076    1     9
28  1076    1     9
29  1078    0     4
30  1078    0     4
31  1078    0     4
32  1078    0     4
33  1078    0     4
34  1078    0     4
35  1121    1     6
36  1121    1     6
37  1121    1     6
38  1121    1     6
39  1121    1     6
40  1121    1     6
41  1161    0     4
42  1161    0     4
43  1161    0     4
44  1161    0     4
45  1161    0     4
46  1161    0     4
47  1209    0     4
48  1209    0     4
49  1209    0     4
50  1209    0     4
...

This is how the output should kind of look like. Industry description = industry

verbal logic:

1) For all unique gvkey create a column which counts the number of fam = 0 in each industry.

2) For all unique gvkey create a column which counts the number of fam = 1 in each industry.

3) Create an output which shows the frequencies of family firms and non family firms for each idnustry

Maybe it even possible to execute this in one code?!

Thank you so much!!

Answer 1

Your verbal logic is not very clear to me (particularly the statements regarding unique gvkey for the final output), but here I provide two results so you can see which one is the thing you want:

• result 1 : using unique(df) for count

dfout <- `colnames<-`(data.frame(as.matrix(aggregate(fam ~industry,
                                                     unique(df),
                                                     FUN = function(x) c(sum(x==0),sum(x==1),sum(x==1)/length(x)*100)))), 
                      c("Industry", "FamCnt", "NoFamCnt", "FamPerc"))

such that

> dfout
  Industry FamCnt NoFamCnt FamPerc
1        4      5        0       0
2        5      1        0       0
3        6      1        1      50
4        9      0        1     100

• result 2 : using df for count

dfout <- `colnames<-`(data.frame(as.matrix(aggregate(fam ~industry,
                                                     df,
                                                     FUN = function(x) c(sum(x==0),sum(x==1),sum(x==1)/length(x)*100)))), 
                      c("Industry", "FamCnt", "NoFamCnt", "FamPerc"))

such that

> dfout
  Industry FamCnt NoFamCnt   FamPerc
1        4     27        0   0.00000
2        5      6        0   0.00000
3        6      5        6  54.54545
4        9      0        6 100.00000

Answer 2

One dplyr otion could be:

df %>%
 group_by(industry) %>%
 summarise(n_family = n_distinct(gvkey[fam == 1]),
           n_no_family = n_distinct(gvkey[fam == 0]),
           perc_family = n_family/n_distinct(gvkey)*100) 

  industry n_family n_no_family perc_family
     <int>    <int>       <int>       <dbl>
1        4        0           5           0
2        5        0           1           0
3        6        1           1          50
4        9        1           0         100

Answer 3

Base R solution (note: it isn't typically good practice to keep spaces in vector names)

# Reshape / Rename the input data: 

ir_df <- setNames(reshape(setNames(aggregate(.~fam+industry, df, length),

                                           c("fam", "industry", "count")),
               direction = "wide",

               idvar = "industry", 

               timevar = "fam"), c("Industry", "Nonfamily Firms", "Family Firms"))

# Transform the data frame to contain the final equation: 

final_df <- transform(replace(ir_df, is.na(ir_df), 0), 

                      `Percent Family Firms In Industry` = 

                        round(`Family Firms` /

                        rowSums(ir_df[,grepl("family", tolower(names(ir_df)))], na.rm = TRUE)

                      * 100, 2))

Data:

df <- structure(list(gvkey = c(1004L, 1004L, 1004L, 1004L, 1004L, 1013L, 
1013L, 1013L, 1013L, 1013L, 1013L, 1045L, 1045L, 1045L, 1045L, 
1045L, 1045L, 1072L, 1072L, 1072L, 1072L, 1072L, 1076L, 1076L, 
1076L, 1076L, 1076L, 1076L, 1078L, 1078L, 1078L, 1078L, 1078L, 
1078L, 1121L, 1121L, 1121L, 1121L, 1121L, 1121L, 1161L, 1161L, 
1161L, 1161L, 1161L, 1161L, 1209L, 1209L, 1209L, 1209L), fam = c(0L, 
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 
0L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 
0L), industry = c(6L, 6L, 6L, 6L, 6L, 4L, 4L, 4L, 4L, 4L, 4L, 
5L, 5L, 5L, 5L, 5L, 5L, 4L, 4L, 4L, 4L, 4L, 9L, 9L, 9L, 9L, 9L, 
9L, 4L, 4L, 4L, 4L, 4L, 4L, 6L, 6L, 6L, 6L, 6L, 6L, 4L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 4L)), class = "data.frame", row.names = c(NA, 
-50L))