Question

我需要转换此表：

1   a   b   0.689723476 0.149916917
    a   b   0.200907662 0.109557062
    a   b   0.684007597 0.703492299
    a   b   0.437375902 0.074223984
    a   b   0.090612241 0.146617232
    a   b   0.526821187 0.895595247
2   a   b   0.507794544 0.731096076
    a   b   0.688987918 0.405801748
    a   b   0.462341505 0.203533346
    a   b   0.750096552 0.889744763
    a   b   0.555074241 0.06610472

进入这个：

1   a   b   0.689723476 0.149916917
1   a   b   0.200907662 0.109557062
1   a   b   0.684007597 0.703492299
1   a   b   0.437375902 0.074223984
1   a   b   0.090612241 0.146617232
1   a   b   0.526821187 0.895595247
2   a   b   0.507794544 0.731096076
2   a   b   0.688987918 0.405801748
2   a   b   0.462341505 0.203533346
2   a   b   0.750096552 0.889744763
2   a   b   0.555074241 0.06610472

基本上，我需要向下重复第一行，直到出现一个新数字，然后重复该数字。我通常使用一种名为Multi-Row Formula的工具在Alteryx中执行此操作，其中我执行快速if语句，如下所示：

if IsNull([Row0:Column1]) then [Row-1:NewColumn] else [Row0:Column1 endif

这将创建一个新列，如果它为null，则取上述单元格中的值，否则它将复制新值。

R或此问题的任何其他解决方案中是否有类似内容？

更新

包含dput数据：

structure(list(X1 = c(1L, NA, NA, NA, NA, NA, 2L, NA, NA, NA, 
NA, NA, NA, NA, NA), X2 = c("a", "a", "a", "a", "a", "a", "a", 
"a", "a", "a", "a", "a", "a", "a", "a"), X3 = c("b", "b", "b", 
"b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b"), 
    X4 = c(0.057520727, 0.54421869, 0.695381681, 0.866708518, 
    0.764246757, 0.622363342, 0.023082188, 0.140441859, 0.404509135, 
    0.806008319, 0.649705949, 0.607341742, 0.275179259, 0.199698605, 
    0.420250037), X5 = c(0.826617034, 0.338437818, 0.069566611, 
    0.95791903, 0.900005669, 0.533003641, 0.075841125, 0.200099759, 
    0.858293828, 0.271342591, 0.308531235, 0.344739272, 0.259006154, 
    0.824994839, 0.610793113)), .Names = c("X1", "X2", "X3", 
"X4", "X5"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-15L), spec = structure(list(cols = structure(list(X1 = structure(list(), class = c("collector_integer", 
"collector")), X2 = structure(list(), class = c("collector_character", 
"collector")), X3 = structure(list(), class = c("collector_character", 
"collector")), X4 = structure(list(), class = c("collector_double", 
"collector")), X5 = structure(list(), class = c("collector_double", 
"collector"))), .Names = c("X1", "X2", "X3", "X4", "X5")), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

Answer 1

我们可以使用fill包中的tidyr函数。 dt是您原始的数据框架。 dt2是最终输出。

library(tidyr)
dt2 <- fill(dt, X1)

在R中复制单元格

1 个答案: