Question

我有一个创建数据框供另一个函数使用的函数。如果我的初始输入是偶数个参数，它会非常有效，但如果它是奇数则会中断。我根本不理解这种行为。

首先，让我们使用两个简单的数据框：three_var和four_var - 变量的数量是指“start”和“mid”变量之间的变量。

three_var
  variable given total
1    start 100.0   100
2        a   0.5    50
3        b   0.7    35
4        c   1.2    42
5      mid    NA    42
6        x   1.0    42
7      end    NA    42

four_var
  variable given total
1    start 100.0 100.0
2        a   0.5  50.0
3        b   1.7  85.0
4        c   0.2  17.0
5        d   1.5  25.5
6      mid    NA  25.5
7        x   1.0  25.5
8      end    NA  25.5

three_var <- structure(list(variable = c("start", "a", "b", "c", "mid", "x", 
"end"), given = c(100, 0.5, 0.7, 1.2, NA, 1, NA), total = c(100, 
50, 35, 42, 42, 42, 42)), .Names = c("variable", "given", "total"
), row.names = c(NA, -7L), class = "data.frame")

four_var <- structure(list(variable = c("start", "a", "b", "c", "d", "mid", 
"x", "end"), given = c(100, 0.5, 1.7, 0.2, 1.5, NA, 1, NA), total = c(100, 
50, 85, 17, 25.5, 25.5, 25.5, 25.5)), .Names = c("variable", 
"given", "total"), row.names = c(NA, -8L), class = "data.frame")

我可以通过以下for子句传递four_var并获得输出。

for (i in 2:nrow(four_var)) {
     four_var$decrease[i] <- four_var$total[i-1] - four_var$total[i]
     four_var$increase[i] <- four_var$total[i] - four_var$total[i-1]
 }
four_var
    variable given total decrease increase
1      start 100.0 100.0       NA       NA
2          a   0.5  50.0     50.0    -50.0
3          b   1.7  85.0    -35.0     35.0
4          c   0.2  17.0     68.0    -68.0
5          d   1.5  25.5     -8.5      8.5
6        mid    NA  25.5      0.0      0.0
7          x   1.0  25.5      0.0      0.0
8        end    NA  25.5      0.0      0.0

然而，当我通过three_var时，我收到错误：

for (i in 2:nrow(three_var)) {
     three_var$decrease[i] <- three_var$total[i-1] - three_var$total[i]
     three_var$increase[i] <- three_var$total[i] - three_var$total[i-1]
 }
Error in `$<-.data.frame`(`*tmp*`, "decrease", value = c(NA, 50)) : 
  replacement has 2 rows, data has 7

此代码看起来不应该取决于输入的数量。我错过了什么？

Answer 1

这不是在data.frame中创建新列的正确方法。它适用于four_var，因为8/2等于4。 i已被回收。您可以验证是否设置了i = 2，运行four_var$decrease[i] <- four_var$total[i-1] - four_var$total[i]并显示four_var。

four_var <- structure(list(variable = c("start", "a", "b", "c", "d", "mid", 
"x", "end"), given = c(100, 0.5, 1.7, 0.2, 1.5, NA, 1, NA), total = c(100, 
50, 85, 17, 25.5, 25.5, 25.5, 25.5)), .Names = c("variable", 
"given", "total"), row.names = c(NA, -8L), class = "data.frame")

i <- 2
four_var$decrease[i] <- four_var$total[i-1] - four_var$total[i]
#   variable given total decrease
# 1    start 100.0 100.0       NA 
# 2        a   0.5  50.0       50
# 3        b   1.7  85.0       NA
# 4        c   0.2  17.0       50 <- recycling
# 5        d   1.5  25.5       NA
# 6      mid    NA  25.5       50 <- recycling
# 7        x   1.0  25.5       NA
# 8      end    NA  25.5       50 <- recycling


three_var$decrease <- NA
three_var$increase <- NA
for (i in 2:nrow(three_var)) {
     three_var$decrease[i] <- three_var$total[i-1] - three_var$total[i]
     three_var$increase[i] <- three_var$total[i] - three_var$total[i-1]
 }
three_var
#   variable given total decrease increase
# 1    start 100.0   100       NA       NA
# 2        a   0.5    50       50      -50
# 3        b   0.7    35       15      -15
# 4        c   1.2    42       -7        7
# 5      mid    NA    42        0        0
# 6        x   1.0    42        0        0
# 7      end    NA    42        0        0

为什么row mutate使用偶数而不是R中的奇数？

1 个答案: