Question

df提供可能的名称匹配。每对匹配应该用空行划分。但是，在某些情况下，我的输出在匹配对之间包含几个空行：

> df <- data.frame(id = c(1,2,NA,3,4,NA,NA,NA,5,6,NA), name = c("john jones", "john joners", 
                   NA, "clara prat", "klara prat", NA, NA, NA, "alan turing", "allan turing", 
                   NA), stringsAsFactors = F)
> df
   id         name
1   1   john jones
2   2  john joners
3  NA         <NA>
4   3   clara prat
5   4   klara prat
6  NA         <NA>
7  NA         <NA>
8  NA         <NA>
9   5  alan turing
10  6 allan turing
11 NA         <NA>

所需的输出是：

> df
   id         name
1   1   john jones
2   2  john joners
3  NA         <NA>
4   3   clara prat
5   4   klara prat
6  NA         <NA>
7   5  alan turing
8   6 allan turing
9  NA         <NA>

我可以使用for循环执行此操作，我理解这不是最佳的。

Answer 1

也许这有帮助

v1 <- rowSums(!is.na(df))
df[unlist(lapply(split(seq_along(v1),
     cumsum(c(1, diff(!v1))<0)), function(i) 
                    i[seq(which.max(v1[i]==0))])),]
#   id         name
#1   1   john jones
#2   2  john joners
#3  NA         <NA>
#4   3   clara prat
#5   4   klara prat
#6  NA         <NA>
#9   5  alan turing
#10  6 allan turing
#11 NA         <NA>

Answer 2

这是另一种使用rle寻找缺失

的方法的方法

miss <- rowSums(is.na(df))

# get runs of missing 
r <- rle(miss)
r$values <- seq_along(r$values)

# subset data, removing rows when all columns are missing 
# and rows sequentially missing
df[!(miss == ncol(df) & duplicated(inverse.rle(r))), ]
#   id         name
# 1   1   john jones
# 2   2  john joners
# 3  NA         <NA>
# 4   3   clara prat
# 5   4   klara prat
# 6  NA         <NA>
# 9   5  alan turing
# 10  6 allan turing
# 11 NA         <NA>

如Akrun所述，您可以使用data.table::rleid来避免某些明确的rle计算

df[!(rowSums(is.na(df)) == ncol(df) & duplicated(data.table::rleid(is.na(df[[1]])))) , ]

Answer 3

使用IRanges包。

df <- data.frame(id = c(1,2,NA,3,4,NA,NA,NA,5,6,NA), name = c("john jones", "john joners", 
                  NA, "clara prat", "klara prat", NA, NA, NA, "alan turing", "allan turing", 
                                         NA), stringsAsFactors = F)

library(IRanges)
na.rs <- which(is.na(df$id) & is.na(df$name))
na.rs.re <- reduce(IRanges(na.rs, na.rs))
na.rs.rm <- na.rs.re[width(na.rs.re)>1]
start(na.rs.rm) <- start(na.rs.rm) + 1

df[-as.integer(na.rs.rm), ]
#    id         name
# 1   1   john jones
# 2   2  john joners
# 3  NA         <NA>
# 4   3   clara prat
# 5   4   klara prat
# 6  NA         <NA>
# 9   5  alan turing
# 10  6 allan turing
# 11 NA         <NA>

Answer 4

当然不是最好的解决方案，但很容易遵循..

miss <- rowSums(is.na(df))
r <- sum(rle(miss)[[2]])

 for(i in 2:length(df$id)){
  while(is.na(df$id[i-1]) & is.na(df$id[i])){
   df <- df[-(i),] 
  if(sum(is.na(df$id)) == r) break
  }
 }

删除R中的连续空行

4 个答案: