df
提出可能的名称匹配。每对匹配项应以空行分隔。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循环来做到这一点,据我所知,这并不是最佳选择。
这是rle
用于寻找遗失的另一种方法
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
避免某些显式rle
计算
df[!(rowSums(is.na(df)) == ncol(df) & duplicated(data.table::rleid(is.na(df[[1]])))) , ]
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