我的数据:
structure(list(col1 = 1:6, col2 = c(NA, NA, 3L, 4L, 5L, 6L),
col3 = c(1L, 1L, NA, 1L, NA, 1L), col4 = c(2L, 2L, 2L, 2L,
NA, 2L), col5 = c(3L, 3L, 3L, NA, 3L, 3L), GROUP = c(1L,
1L, 1L, 2L, 2L, 2L)), class = "data.frame", row.names = c(NA, -6L))
我想得到一个新的日期框,如果出现NA,那么用组合list(c(1,2),c(3,4,5))
中的NA替换整行.
我想要的是:
data= structure(list(col1 = c(NA, NA, 3L, 4L, 5L, 6L), col2 = c(NA,
NA, 3L, 4L, 5L, 6L), col3 = c(1L, 1L, NA, NA, NA, 1L), col4 = c(2L,
2L, NA, NA, NA, 2L), col5 = c(3L, 3L, NA, NA, NA, 3L), GROUP = c(1L,
1L, 1L, 2L, 2L, 2L)), class = "data.frame", row.names = c(NA,
-6L))
更新:
list_v <- list(c(1,2),c(3,4,5))
data1 <- do.call(qpcR:::cbind.na, lapply(list_v, function(i)data[complete.cases(data[c(i)]),i]))
我不需要删除至少有一个NA的行,而是用至少有一个NA的NA替换其余元素