Question

我需要一些帮助。我有下表：

country_code=c(1,1,1,1,1,1,2,2,2,2,2,2)
target=c('V1','V1','V2','V2','V3','V3','V1','V1','V2','V2','V3','V3')
M1=c('X7','X7','X14','X14','X8','X8','X29','X22','X2','X22','X22','X22')
M2=c('X1','X1','X17','X11','X21','X21','X1','X29','X8','X18','X24','X24')
M3=c('NA','NA','NA','X1','NA','NA','NA','NA','NA','NA','NA','NA')
CountofRun=c(1,2,1,2,1,2,1,2,1,2,1,2)
df<-data.frame(country_code,target,M1,M2,M3,CountofRun)

我希望获得每个 country_code 和目标组合的频率表。因此，例如，如果在 country_code = 1 和 target = V1 的所有三次运行中出现 X7 ，则需要将X7求和为3.将会看到，我只想计算每个X1到X30在country_code和target的6种组合中的每一种中出现的次数。我无法转换为数字。

终极表，希望看起来像这样

Answer 1

也许

library(dplyr)
library(tidyr)

df %>%
  select(-CountofRun) %>%
  gather(key, value, -(country_code:target)) %>%
  select(-key) %>%
  ftable(xtabs(~ country_code + target + value, data = .))

给出了：

#                    value NA X1 X11 X14 X17 X18 X2 X21 X22 X24 X29 X7 X8
#country_code target                                                     
#1            V1            2  2   0   0   0   0  0   0   0   0   0  2  0
#             V2            1  1   1   2   1   0  0   0   0   0   0  0  0
#             V3            2  0   0   0   0   0  0   2   0   0   0  0  2
#2            V1            2  1   0   0   0   0  0   0   1   0   2  0  0
#             V2            2  0   0   0   0   1  1   0   1   0   0  0  1
#             V3            2  0   0   0   0   0  0   0   2   2   0  0  0

Answer 2

data.table解决方案（与dplyr + tidyr的结构类似，只是语法不同）

setDT(df)
df[, .SD
   ][, CountofRun := NULL
   ][, melt(.SD, id.vars=c('country_code', 'target'))
   ][, .N, .(country_code, target, value)
   ][, dcast(.SD, country_code + target ~ value, value.var='N', fill=0)
   ]

Answer 3

这会让你分道扬;你有计数现在它只是格式化：

> library(data.table)
> 
> country_code=c(1,1,1,1,1,1,2,2,2,2,2,2)
> target=c('V1','V1','V2','V2','V3','V3','V1','V1','V2','V2','V3','V3')
> M1=c('X7','X7','X14','X14','X8','X8','X29','X22','X2','X22','X22','X22')
> M2=c('X1','X1','X17','X11','X21','X21','X1','X29','X8','X18','X24','X24')
> M3=c('NA','NA','NA','X1','NA','NA','NA','NA','NA','NA','NA','NA')
> CountofRun=c(1,2,1,2,1,2,1,2,1,2,1,2)
> df<-data.table(country_code,target,M1,M2,M3,CountofRun)
> 
> # melt the data for easier processing
> df_m <- melt(df, id.vars = c('country_code', 'target', 'CountofRun'))
> 
> # count
> df_count <- df_m[, 
+             .(count = sum(CountofRun)),
+             keyby = .(country_code, target, value)
+             ][value != "NA"]  # remove 'NA's
>             
> df_count
    country_code target value count
 1:            1     V1    X1     3
 2:            1     V1    X7     3
 3:            1     V2    X1     2
 4:            1     V2   X11     2
 5:            1     V2   X14     3
 6:            1     V2   X17     1
 7:            1     V3   X21     3
 8:            1     V3    X8     3
 9:            2     V1    X1     1
10:            2     V1   X22     2
11:            2     V1   X29     3
12:            2     V2   X18     2
13:            2     V2    X2     1
14:            2     V2   X22     2
15:            2     V2    X8     1
16:            2     V3   X22     3
17:            2     V3   X24     3
>

R由两个因子计算的字符串数

3 个答案: