Question

我试图读取一些文件，其中某些行在文本字符串中包含额外的分号（我不知道是什么导致了这种情况）

作为一个例子，这是一个具有相同问题的超简化数据：

bad_data <- "100; Mc Donalds; Seattle; normal day
             115; Starbucks; Boston; normal day
             400; PF Chang; Chicago; busy day
             400;; Texas; busy day
             10; D;unkin Donuts; Washin;gton; lazy day"

所以它没有标题，我试着用它来阅读：

library(data.table)
fread(bad_data, sep = ";", header = F, na.strings = c("", NA), strip.white = T)

但是没有雪茄......这有点不可能阅读，如果没有干净的解决方案，我想跳过这些行。

Answer 1

如果您只想删除没有预期分隔符数的行：

#get all Animal_IDs from capture dataset
allID = unique(capdat$Animal_ID)

#create list to hold data frames, one df for each animalID
df.list <- as.list(rep("", length(allID)))

#loop through each animal ID, find matching collar serial #, capture date, 
#and mortality date (if applicable)
for (i in 1:length(allID)){
  ID.i = allID[i]
  ser.i <- pull(capdat[capdat$Animal_ID == ID.i, 4])
  capdate.i = pull(capdat[capdat$Animal_ID == ID.i, 2])
  mortdate.i = pull(capdat[capdat$Animal_ID == ID.i, 11])

  ifelse(is.na(mortdate.i), 
        df.list[[i]] <- dat[(dat$CollarSerialNumber == ser.i & 
                          dat$Date > capdate.i) ,],
        df.list[[i]] <- dat[(dat$CollarSerialNumber == ser.i & 
                          dat$Date > capdate.i & dat$Date < mortdate.i) ,])
  df.list[[i]]$Animal_ID = ID.i
  }

#merge list into a single data frame
df <- ldply(df.list, data.frame)

结果：

library(stringi)
library(magrittr)

bad_data <- 
"100; Mc Donalds; Seattle; normal day
115; Starbucks; Boston; normal day
400; PF Chang; Chicago; busy day
400;; Texas; busy day
10; D;unkin Donuts; Washin;gton; lazy day"

# split to lines. you could also use readLines if it's coming from a file
text_lines <- unlist(strsplit(bad_data, '\n'))

# which lines contain the expected number of semicolons?
good_lines <- sapply(text_lines, function(x) stri_count_fixed(x, ';') == 3)

# for those lines, split to vectors and (optional bonus) trim whitespace
good_vectors <- lapply(
  text_lines[good_lines], 
  function(x) x %>% strsplit(';') %>% unlist %>% trimws)

# flatten to matrix (from which you can make a data.frame or whatever you want)
my_mat <- do.call(rbind, good_vectors)

Answer 2

您可以尝试删除文本字符串中的所有分号（这假设所有不需要的分号都完全在字符串中：

gsub("(\\S);(\\S)", "\\1\\2", bad_data, perl=TRUE)
[1] "100; Mc Donalds; Seattle; normal day\n             115; Starbucks; Boston; normal day\n             400; PF Chang; Chicago; busy day\n             400; Texas; busy day\n             10; Dunkin Donuts; Washington; lazy day"

读取分隔文件，其中分号显示为分隔符和字符串

2 个答案: