自动填充 r 中的列答案

【问题标题】：Automated fill in columns in r自动填充 r 中的列
【发布时间】：2021-11-05 11:28:45
【问题描述】：

我有一个数据框（如下所示），其中“sig”列中有一些星号。

我想在有星号的最下方行上方的 sig 列的空单元格中填写星号，在这种情况下，从“H”行开始，星号将无处不在，如下所示：

我正在考虑某种 for 循环，它可以识别最下面的行，其中有一个星号，然后在上面的空单元格中填充星号可能是要走的路，但我不知道如何编码这个。

出于调试目的，我在 R 中使用

df<- data.frame("variable"= c("a","b","c","d","e","f","g","h","i","j","k"),
                "value" = c(0.04,0.03,0.04,0.02,0.03,0.02,0.02,0.01,0.04,0.1,0.02), 
                "sig" = c("*","*","*","","*","*","","*","","",""))

任何帮助将不胜感激 - 谢谢！

【问题讨论】：

标签： r dataframe for-loop

【解决方案1】：

我们可以根据找到具有*的最后一个元素的索引来使用replace

library(dplyr)
df <- df %>%
    mutate(sig = replace(sig, seq(tail(which(sig == "*"), 1)), "*"))

-输出

df
   variable value sig
1         a  0.04   *
2         b  0.03   *
3         c  0.04   *
4         d  0.02   *
5         e  0.03   *
6         f  0.02   *
7         g  0.02   *
8         h  0.01   *
9         i  0.04    
10        j  0.10    
11        k  0.02

【讨论】：

【解决方案2】：

另一种方式：

df[1:max(which(df$sig == "*")), "sig"] = "*"

给予：

   variable value sig
1         a  0.04   *
2         b  0.03   *
3         c  0.04   *
4         d  0.02   *
5         e  0.03   *
6         f  0.02   *
7         g  0.02   *
8         h  0.01   *
9         i  0.04    
10        j  0.10    
11        k  0.02

【讨论】：

【解决方案3】：

另一种解决方案是使用fill，但您需要将"" 更改为NA

库

library(tidyverse)

数据

df <-
  data.frame("variable"= c("a","b","c","d","e","f","g","h","i","j","k"),
             "value" = c(0.04,0.03,0.04,0.02,0.03,0.02,0.02,0.01,0.04,0.1,0.02), 
             "sig" = c("*","*","*","","*","*","","*","","",""))

代码

df %>% 
  mutate(sig = if_else(sig == "",NA_character_,sig)) %>% 
  fill(sig,.direction = "up")

输出

   variable value  sig
1         a  0.04    *
2         b  0.03    *
3         c  0.04    *
4         d  0.02    *
5         e  0.03    *
6         f  0.02    *
7         g  0.02    *
8         h  0.01    *
9         i  0.04 <NA>
10        j  0.10 <NA>
11        k  0.02 <NA>

【讨论】：