使用Rowname作为标识符替换多个值

Question

我一直试图将列中的多个值替换为它们的行名。在我的实际数据中，它们代表了一种主观测试，在这一点上，它要么被记录为阳性，要么记录为阴性。我必须在我的数据框架中对它进行重新分类，我所要做的就是示例ID，即行名。

与其进入并手动更改每个特定值，我想知道是否有一种方法可以同时执行多个值。我看了一下this question。我试过这个，

        dat <- structure(list(A = structure(c(2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L,2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L), .Label = c("1", "0"), class = "factor"),B = structure(c(1L,1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,1L, 1L, 1L, 1L, 2L, 1L, 1L), .Label = c("0", "1"), class = "factor"),C = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L), .Label = c("nd","0", "1"), class = "factor"),D = structure(c(1L, 1L, 1L,2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 2L,2L, 1L, 1L, 1L, 1L, 1L), .Label = c("0", "1"), class = "factor"),E = structure(c(1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("0","1"), class = "factor")),.Names = c("A", "B", "C", "D", "E"), class = "data.frame", row.names = c(NA, 24L))

        dat$result <- as.integer(rowSums(dat[,1:5] == "1")> 0)
        dn <- c("1","5","7","10","14","15","16")
        dat$result[dn] <- "3"

有人能帮我处理这个吗。

Answer 1

dn必须是数字，而不是字符。

当您以dn作为字符运行命令时，您将只获得NA值：

dat$result[dn]
# [1] NA NA NA NA NA NA NA

如果将dn更改为数字，将得到正确的值：

dat$result[as.numeric(dn)]
# [1] 0 1 0 1 0 1 0

然后您可以指定如下的新值：

dat$result[as.numeric(dn)] <- 3
dat$result
# [1] 3 0 1 1 3 1 3 1 0 3 0 1 1 3 3 3 1 1 1 0 0 1 0 1

这将不会由data.frame通过row.names过滤，而是按识别码过滤，但是由于它是一个有序的序列，所以您可以只使用这些索引进行转换。还是需要基于row.names进行匹配？

要通过row.names进行过滤，您可以执行以下操作：

## Filter by rownames
row.names(dat) <- paste0("row_", row.names(dat))
dat

dn <- c("row_1","row_5","row_7")

dat[row.names(dat) %in% dn,]$result <- 3

dat