有條件指望基於

Question

列

之間的比較值的行，我有以下的數據幀：

df <- structure(list(x = c(0.389794300700167, -1.20807617542949, -0.363676017470862, 
-1.62667268170309, -0.256478394123992, 1.10177950308713, 0.755781508027337, 
-0.238233556018718, 0.98744470341339, 0.741390128383824), y = c(0.0893472664958216, 
-0.954943856152377, -0.195150384667239, 0.92552126209408, 0.482978524836611, 
-0.596310636720207, -2.18528683816953, -0.674865937875116, -2.11906119191017, 
-1.2651980215309), fac = structure(c(2L, 1L, 2L, 3L, 1L, 1L, 
1L, 1L, 2L, 2L), .Label = c("A", "B", "C"), class = "factor")), .Names = c("x", 
"y", "fac"), row.names = c(NA, -10L), class = "data.frame") 


df 
#>    x   y fac  manual_assignment 
#> 1 0.3897943 0.08934727 B  b.x 
#> 2 -1.2080762 -0.95494386 A  a.y 
#> 3 -0.3636760 -0.19515038 B  b.y 
#> 4 -1.6266727 0.92552126 C  c.y 
#> 5 -0.2564784 0.48297852 A  a.y 
#> 6 1.1017795 -0.59631064 A  a.x 
#> 7 0.7557815 -2.18528684 A  a.x 
#> 8 -0.2382336 -0.67486594 A  a.x 
#> 9 0.9874447 -2.11906119 B  b.x 
#> 10 0.7413901 -1.26519802 B  b.x 

我想要做的是計算基於在x和y比較值的行。對於每一行，如果該值x比y較大，我們增加fac成員的數量由1所以最後的結果是這樣的：

 x.count y.count 
A  3   2 (# a.y) 
B  3   1 
C  0   1 

我怎樣才能做到這一點？可能與dplyr？

Answer 1

這是更直接的與table：

with(df, table(fac, ifelse(x > y, "x.count", "y.count")))  

#fac x.count y.count 
# A  3  2 
# B  3  1 
# C  0  1 

---

隨着dplyr/tidyr，你需要的代碼幾行：

library(tidyverse) 
df %>% 
    group_by(fac, measure = if_else(x > y, "x.count", "y.count")) %>% 
    tally() %>% 
    spread(measure, n, fill = 0) 

#Source: local data frame [3 x 3] 
#Groups: fac [3] 

#  fac x.count y.count 
#* <fctr> <dbl> <dbl> 
#1  A  3  2 
#2  B  3  1 
#3  C  0  1 

Answer 2

隨着dplyr我們可以按fac和計數其中x和y值更大的值的length。

library(dplyr) 
df %>% 
    group_by(fac) %>% 
    summarise(x.count = length(which(x > y)), 
      y.count = length(which(x < y))) 


#  fac x.count y.count 
# <fctr> <int> <int> 
#1  A  3  2 
#2  B  3  1 
#3  C  0  1 

Answer 3

隨着data.table，我們可以使用

library(data.table) 
dcast(setDT(df)[, .N, .(fac, measure = c('y.count', 'x.count')[(x > y) + 1])], 
      fac ~measure, fill = 0) 
# fac x.count y.count 
#1: A  3  2 
#2: B  3  1 
#3: C  0  1