Question

我有一个data.frame，它有几个零值的变量。我需要构造一个额外的变量，它将返回每个观察值不为零的变量组合。 E.g。

df <- data.frame(firm = c("firm1", "firm2", "firm3", "firm4", "firm5"),
                 A = c(0, 0, 0, 1, 2),
                 B = c(0, 1, 0, 42, 0),
                 C = c(1, 1, 0, 0, 0))

现在我想生成新变量：

df$varCombination <- c("C", "B-C", NA, "A-B", "A")

我想到了这样的东西，这显然不起作用：

for (i in 1:nrow(df)){
    df$varCombination[i] <- paste(names(df[i,2:ncol(df) & > 0]), collapse = "-")
}

Answer 1

这可能很容易使用apply(df, 1, fun)来解决，但是为了性能起见，这里尝试解决这个列而不是行（我曾经看过@alexis_laz做过的类似但却无法找到它现在）

## Create a logical matrix
tmp <- df[-1] != 0
## or tmp <- sapply(df[-1], `!=`, 0)

## Prealocate result 
res <- rep(NA, nrow(tmp))

## Run per column instead of per row
for(j in colnames(tmp)){
  res[tmp[, j]] <- paste(res[tmp[, j]], j, sep = "-")
}

## Remove the pre-allocated `NA` values from non-NA entries
gsub("NA-", "", res, fixed = TRUE)
# [1] "C"   "B-C" NA    "A-B" "A"

更大数据集的一些基准

set.seed(123)
BigDF <- as.data.frame(matrix(sample(0:1, 1e4, replace = TRUE), ncol = 10))

library(microbenchmark)

MM <- function(df) {
  var_names <- names(df)[-1]
  res <- character(nrow(df))
  for (i in 1:nrow(df)){
    non_zero_names <- var_names[df[i, -1] > 0]
    res[i] <- paste(non_zero_names, collapse  = '-')
  }
  res
}

ZX <- function(df) {
  res <- 
    apply(df[,2:ncol(df)]>0, 1,
          function(i)paste(colnames(df[, 2:ncol(df)])[i], collapse = "-"))
  res[res == ""] <- NA
  res
}

DA <- function(df) {
  tmp <- df[-1] != 0
  res <- rep(NA, nrow(tmp))

  for(j in colnames(tmp)){
    res[tmp[, j]] <- paste(res[tmp[, j]], j, sep = "-")
  }
  gsub("NA-", "", res, fixed = TRUE)
}


microbenchmark(MM(BigDF), ZX(BigDF), DA(BigDF))
# Unit: milliseconds
#      expr       min         lq       mean     median         uq        max neval cld
# MM(BigDF) 239.36704 248.737408 253.159460 252.177439 255.144048 289.340528   100   c
# ZX(BigDF)  35.83482  37.617473  38.295425  38.022897  38.357285  76.619853   100  b 
# DA(BigDF)   1.62682   1.662979   1.734723   1.735296   1.761695   2.725659   100 a

Answer 2

使用apply：

# paste column names
df$varCombination <- 
  apply(df[,2:ncol(df)]>0, 1,
        function(i)paste(colnames(df[, 2:ncol(df)])[i], collapse = "-"))

# convert blank to NA
df$varCombination[df$varCombination == ""] <- NA

# result
df
#    firm A  B C varCombination
# 1 firm1 0  0 1              C
# 2 firm2 0  1 1            B-C
# 3 firm3 0  0 0           <NA>
# 4 firm4 1 42 0            A-B
# 5 firm5 2  0 0              A

Answer 3

You had the right idea but the logical comparison in your loop wasn't correct.

I've attempted to keep the code fairly similar to what you had before, this should work:

var_names <- names(df)[-1]

df$varCombination <- character(nrow(df))

for (i in 1:nrow(df)){

  non_zero_names <- var_names[df[i, -1] > 0]

  df$varCombination[i] <- paste(non_zero_names, collapse  = '-')

}

> df
   firm A  B C varCombination
1 firm1 0  0 1              C
2 firm2 0  1 1            B-C
3 firm3 0  0 0               
4 firm4 1 42 0            A-B
5 firm5 2  0 0              A

如何在R data.frame中创建组合变量？

3 个答案: