R 函数中的相对引用观察

Question

在编写一个函数来计算向量中的每个观察值时，我如何引用所述观察值以包括距离当前正在操作的观察值有预定数量的观察值的观察值单元格？如果每一行都是 i，例如 i = 1、2、... 等，我如何引用第 i-1 行中的列？

这是一个模拟我的困境的示例数据集：

> letters <- c('a', 'b', 'c', 'b', 'e')
> numbers <- c('1', '', '2', '', '3')
> sample <- cbind(letters, numbers)
> sample
     letters numbers
[1,] "a"     "1"    
[2,] "b"     ""     
[3,] "c"     "2"    
[4,] "b"     ""     
[5,] "e"     "3"  

我想用之前观察到的

sample$numbers

中的值填充

sample$numbers

中的每个空单元格。我如何引用在其创建过程中创建的观察？例如，我试过：

> sample$numbers <- ifelse(sample$numbers == "", sample$numbers[as.numeric(rownames(sample)) - 1], sample$numbers)
Error in sample$numbers : $ operator is invalid for atomic vectors

我也试过在

b

中使用常见的

sample$letters

来填充缺失值：

> f1 <- function(df, cols, match_with, to_x = 'b'){
+   df[cols] <- lapply(df[cols], function(i) 
+     ifelse(grepl(to_x, match_with, fixed = TRUE), sample$numbers[as.numeric(rownames(sample)) - 1], 
+            i))
+   return(df)
+ }
> sample = f1(sample, cols = c('numbers'), match_with = sample$letters)
 Hide Traceback

 Rerun with Debug
 Error in sample$letters : $ operator is invalid for atomic vectors 
5.
grepl(to_x, match_with, fixed = TRUE) 
4.
ifelse(grepl(to_x, match_with, fixed = TRUE), sample$numbers[as.numeric(rownames(sample)) - 
    1], i) 
3.
FUN(X[[i]], ...) 
2.
lapply(df[cols], function(i) ifelse(grepl(to_x, match_with, fixed = TRUE), 
    sample$numbers[as.numeric(rownames(sample)) - 1], i)) 
1.
f1(sample, cols = c("numbers"), match_with = sample$letters) 

我的麻烦似乎是，在这两种情况下，我都在使用

sample$numbers[as.numeric(rownames(sample)) - 1]

来引用

sample$numbers

在先前观察中的值。有更好的方法吗？

Answer 1

sample[,"numbers"] <- sapply(seq_along(sample[,"numbers"]), 
                             function(x) ifelse(sample[,"numbers"][x] == '', 
                                                sample[,"numbers"][x-1], 
                                                sample[,"numbers"][x]))

     letters numbers
[1,] "a"     "1"    
[2,] "b"     "1"    
[3,] "c"     "2"    
[4,] "b"     "2"    
[5,] "e"     "3"    

Answer 2

假设您有一个 data.frame 而不是上面使用的矩阵（为了能够使用

$

引用列），您可以为此使用

zoo::na.locf

：

#make a data.frame instead of a matrix
sample <- data.frame(letters, numbers)

library(zoo)
#if your data has '' empty cells then convert those to NA
sample$numbers[sample$numbers == ''] <- NA
sample$numbers <- na.locf(sample$numbers)

输出：

sample
  letters numbers
1       a       1
2       b       1
3       c       2
4       b       2
5       e       3

Answer 3

您可以使用

FillDown

包中的

DataCombine

功能：

library(DataCombine)
letters <- c('a', 'b', 'c', 'b', 'e')
numbers <- c('1', '', '2', '', '3')
numbers[numbers==""] <- NA # replace empty strings with NA
sample <- data.frame(letters,numbers)

FillDown(sample,"numbers")

Answer 4

letters <- c('a', 'b', 'c', 'b', 'e')
numbers <- c('1', '', '2', '', '3')
sample <- data.frame(letters, numbers, stringsAsFactors = F)


sample$numbers[sample$numbers == ""] <- c(sample$numbers[2:nrow(sample)], NA)[sample$numbers == ""]