如果我创建一个数据框
df = data.frame(a=c(1,2,3), b=c(4,5,6))
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为什么这段代码
df$z[c(1,2)] = c(7,8)
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产生这个错误
Error in `$<-.data.frame`(`*tmp*`, z, value = c(7, 8)) :
replacement has 2 rows, data has 3
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这段代码有效吗?
df$z[c(2,3)] = c(7,8)
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df
a b z
1 1 4 NA
2 2 5 7
3 3 6 8
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如果赋值是一个向量,R 似乎会创建z要添加到 的向量data.frame,并且该向量需要与 中的数字行具有相同的长度data.frame。
如果您使用对象,会更清楚地显示发生的情况list:
df <- list(a = 1:3, b = 4:6)
df$z1[1:2] <- 7:8
df$z2[2:3] <- 7:8
df$z3[c(1,3)] <- 7:8
df
#> $a
#> [1] 1 2 3
#>
#> $b
#> [1] 4 5 6
#>
#> $z1
#> [1] 7 8
#>
#> $z2
#> [1] NA 7 8
#>
#> $z3
#> [1] 7 NA 8
data.frame(df)
#> Error in (function (..., row.names = NULL, check.rows = FALSE, check.names = TRUE, : arguments imply differing number of rows: 3, 2
data.frame(df[-3])
#> a b z2 z3
#> 1 1 4 NA 7
#> 2 2 5 7 NA
#> 3 3 6 8 8
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错误来自$<-.data.frame函数
> `$<-.data.frame`
function (x, name, value)
{
cl <- oldClass(x)
class(x) <- NULL
nrows <- .row_names_info(x, 2L)
if (!is.null(value)) {
N <- NROW(value)
if (N > nrows)
stop(sprintf(ngettext(N, "replacement has %d row, data has %d",
"replacement has %d rows, data has %d"), N, nrows),
domain = NA)
...
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N > nrows即满足 的条件
> NROW(c(7, 8))
[1] 2
> .row_names_info(df, 2L)
[1] 3
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这是通过traceback()错误确认的
> traceback()
3: stop(sprintf(ngettext(N, "replacement has %d row, data has %d",
"replacement has %d rows, data has %d"), N, nrows), domain = NA)
2: `$<-.data.frame`(`*tmp*`, z, value = c(7, 8))
1: `$<-`(`*tmp*`, z, value = c(7, 8))
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