如何提取多个 zip 文件并在 R 中读取这些 csvs? [英] How can I extract multiple zip files and read those csvs in R?

问题描述

我需要一次解压缩多个文件并将其作为数据框添加到我的 R 仪表板中.

I need to unzip multiple files at once and add it as a data frame in my R dashboard.

我目前正在使用此代码:

I'm currently using this code:

zipF<- "/Users/sahilverma13/Desktop/chat_data_2017-01-30_IST.zip"
outDir<-"/Users/sahilverma13/Desktop"
unzip(zipF,exdir=outDir)

但我必须为每个文件单独做.

But I have to do it for every file separately.

zipF <- list.files(pattern="*.zip")

我尝试使用通配符，但它不起作用.

I tried using the wildcard but it doesn't work.

请帮忙.

推荐答案

我经常使用 plyr 包中的 ldply 函数来读取或处理多个文件.

I often use the ldply function from the plyr package to read or do stuff with multiple files.

library(plyr)

# get all the zip files
zipF <- list.files(path = "/your/path/here/", pattern = "*.zip", full.names = TRUE)

# unzip all your files
ldply(.data = zipF, .fun = unzip, exdir = outDir)

正如理查德指出的那样，这并不完整(早上喝太多咖啡也不好).

As Richard pointed out this is not complete (to much coffee in the morning is also not good).

# get the csv files
csv_files <- list.files(path = outDir, pattern = "*.csv")

# read the csv files
my_data <- ldply(.data = csv_files, .fun = read.csv)

我非常喜欢乔尔的评论.太习惯用plyr包了，忘了你也可以用sapply函数.也许更好用！

I liked the comment of Joel a lot. I'am used to using the plyr package so much that I forgot that you can also use the sapply function. Maybe even better to use!

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