1

我在 R 中有以下数据框：

df <- data.frame(id=c('a','b','a','c','b','a'),
                 indicator1=c(1,0,0,0,1,1),
                 indicator2=c(0,0,0,1,0,1),
                 extra1=c(4,5,12,4,3,7),
                 extra2=c('z','z','x','y','x','x'))

id indicator1 indicator2 extra1 extra2
a          1          0      4      z
b          0          0      5      z
a          0          0     12      x
c          0          1      4      y
b          1          0      3      x
a          1          1      7      x

我想添加一个新列，该列对特定 id 出现各种指标等于 1 的次数的所有行进行计数。例如：

id indicator1 indicator2 extra1 extra2 countInd1 countInd2 countInd1Ind2
a          1          0      4      z      2         1           1
b          0          0      5      z      1         0           0
a          0          0     12      x      2         1           1
c          0          1      4      y      0         1           0
b          1          0      3      x      1         0           0
a          1          1      7      x      2         1           1

我怎样才能做到这一点？

4

2 回答 2

4

有几种方法。这是一个ave和within：

within(df, {
  ind1ind2 <- ave(as.character(interaction(indicator1, indicator2, drop=TRUE)), 
                  id, FUN = function(x) sum(x == "1.1"))
  ind2 <- ave(indicator2, id, FUN = function(x) sum(x == 1))
  ind1 <- ave(indicator1, id, FUN = function(x) sum(x == 1))
})
#   id indicator1 indicator2 extra1 extra2 ind1 ind2 ind1ind2
# 1  a          1          0      4      z    2    1        1
# 2  b          0          0      5      z    1    0        0
# 3  a          0          0     12      x    2    1        1
# 4  c          0          1      4      y    0    1        0
# 5  b          1          0      3      x    1    0        0
# 6  a          1          1      7      x    2    1        1

这是一个替代方案：

A <- setNames(aggregate(cbind(indicator1, indicator2) ~ id, df, 
                        function(x) sum(x == 1)), c("id", "ind1", "ind2"))
B <- setNames(aggregate(interaction(indicator1, indicator2, drop = TRUE) ~ id, 
                        df, function(x) sum(x == "1.1")), c("id", "ind1ind2"))
Reduce(function(x, y) merge(x, y), list(df, A, B))
#   id indicator1 indicator2 extra1 extra2 ind1 ind2 ind1ind2
# 1  a          1          0      4      z    2    1        1
# 2  a          0          0     12      x    2    1        1
# 3  a          1          1      7      x    2    1        1
# 4  b          0          0      5      z    1    0        0
# 5  b          1          0      3      x    1    0        0
# 6  c          0          1      4      y    0    1        0

当然，如果你的数据很大，你会想要探索“data.table”包。within它也比版本少一点打字。

library(data.table)
DT <- data.table(df)
DT[, c("ind1", "ind2", "ind1ind2") := 
     list(sum(indicator1 == 1),
          sum(indicator2 == 1),
          sum(interaction(indicator1, indicator2, 
                          drop = TRUE) == "1.1")),
   by = "id"]
DT
#    id indicator1 indicator2 extra1 extra2 ind1 ind2 ind1ind2
# 1:  a          1          0      4      z    2    1        1
# 2:  b          0          0      5      z    1    0        0
# 3:  a          0          0     12      x    2    1        1
# 4:  c          0          1      4      y    0    1        0
# 5:  b          1          0      3      x    1    0        0
# 6:  a          1          1      7      x    2    1        1

而且，如果您觉得这更明确sum(interaction(...) == "1.1")，您也可以这样做。sum(indicator1 == 1 & indicator2 == 1)我还没有进行基准测试，看看哪个更有效。interaction这正是我首先想到的。

于 2013-08-20T17:07:06.233 回答
0

或者你可以这样做：

get_freq1 = function(i) {sum(df[which(df$id == df[i,1]),]$indicator1)}
get_freq2 = function(i) {sum(df[which(df$id == df[i,1]),]$indicator2)} 

df = data.frame(df, countInd1 = sapply(1:nrow(df), get_freq1), countInd2 = sapply(1:nrow(df), get_freq2))
df= data.frame(df, countInd1Ind2 = ((df$countInd1 != 0) & (df$countInd2 != 0))*1)

你得到：

 #  id indicator1 indicator2 extra1 extra2 countInd1 countInd2 countInd1Ind2
 #1  a          1          0      4      z         2         1             1
 #2  b          0          0      5      z         1         0             0
 #3  a          0          0     12      x         2         1             1
 #4  c          0          1      4      y         0         1             0
 #5  b          1          0      3      x         1         0             0
 #6  a          1          1      7      x         2         1             1
于 2013-08-20T17:21:24.417 回答