Question

我有24个数据文件（bsls）。每个文件包含固定数量的行，但列数可变（sites）。我有一个清晰的列表23 sites但不能完全匹配，因为与每个站点相关的列名包含其他信息。

我已使用以下代码将这些文件读入R：

#list files from dir and read, skipping rows until 'Q Num'
temp <- list.files() # e.g. info-stuff-nameofbsl-otherStuff.csv

# read.xls and strip bsl name from file and assign as object name
for(i in temp){
    assign(unlist(strsplit(i, split = '-', fixed = T))[3],
           read.xls(i, pattern = "Q Num"))
}

#create list of dataframes (24 bsls)
bsls <- Filter(function(x) is(x, "data.frame"), mget(ls()))

#clean list of site names
sites <- ("NewYork","London","Sydney","Paris","Manchester","Angers","Venice","Bangkok","Glasgow","Boston","Perth","Canberra","Lyons","Washington","Milan","Cardiff","Dublin","Frankfurt","Ottawa","Toronto","El.Salvador","Taltal","Caldera")

24个bsls数据集中的1个的前3行示例

e.g。 BSL1

QNum,   QuestionText,   % unrelatedCol, NewYork_Other_info, London_some_other_info, Venice_other_diff_info, 
q17a,   question?,                 74%,              69%,                     81%,                  76%,
q17b,   Another question?,         72%,              73%,                     77%,                  74%,

我需要的结果是，23个sites中的每一个都有一个.csv文件，其中包含24个数据文件（bsls）中找到的所有列。

我目前的尝试......

for(site in sites){                             #for each site
    assign(site, data.frame())                  #create empty data frame to add vectors to
    for(bsl in dfs){                            #for each dataset
        if (grepl(site, colnames(bsl))){        #substring match
           next                                 #go back to for loop
        }
    assign(site$bsl, bsl[,grepl("site", colnames(bsl))]) #assign column to dataframe
    } 
}

解决方案看起来像这样......

例如London.csv

QNum,   QuestionText, BSLname1_Other_info,  BSLname2_some_other_info, BSL5other_diff_info, 
q17a,   question?,                 74%,              69%,                     81%,                  76%,
q17b,   Another question?,         72%,              73%,                     77%,                  74%,

将有23个文件，每个站点一个，包含24个输入bsl文件中与该站点相关的列。

编辑 - 值得说明bsls中的每一个都不会被称为bsl1，bsl2 ...等等，但实际上是唯一的字符串，例如unit，section，team等等。

Answer 1

library(dplyr)
library(stringi)
library(tidyr)


bind_rows(bsls, .id = bsl) %>%
  gather(variable, value, 
         matches(sites %>% paste(collapse = "|") ), 
         na.rm = TRUE ) %>%
  separate(variable, c("site", "new_variable", 
           sep = "_", extra = "merge") %>%
  unite(final_variable, bsl, new_variable, sep = "_") %>%
  spread(final_variable, value) %>%
  group_by(site) %>%
  do(write.csv(., paste("site", first(.$site), ".csv") ) )

Answer 2

以下代码最终解决了我的问题。我首先必须通过重命名bsls之前的for loop数据框列表中的所有列来进一步打破原始问题。这是为了知道bsl所属的site - 可以找到重命名逻辑here。

循环解决方案

#this loop prints the files
for (site in sites){
    #create new file with question cols only
    newfile <- data.frame(NewYork[,1:2], stringsAsFactors = F)
    # search for columns in bsls relating to site
    for (bsl in bsls){
        colids <- grepl(site, colnames(bsl))
        cols <- bsl[,colids, drop = F]
        newfile <- cbind(newfile, cols)
        }
    filename <- paste0("Site ", site," .csv")
    write.xlsx(newfile, file = filename, row.names = F)
}

如何循环遍历多个数据帧，查找字符串列表，搜索包含字符串的列并创建多个新文件？

2 个答案: