blocks|key|1221189|text|为了记录，我试着把Dirk的答案翻译成代码:-P|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1221190|temp+<-+tempfile()
download.file("http://www.newcl.org/data/zipfiles/a1.zip",temp)
con+<-+unz(temp,+"a1.dat")
data+<-+matrix(scan(con),ncol=4,byrow=TRUE)
unlink(temp)|code-block|syntax|javascript|1221191|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Just for the record, I tried translating Dirk's answer into code :-P

<pre><code>temp &lt;- tempfile()
download.file("http://www.newcl.org/data/zipfiles/a1.zip",temp)
con &lt;- unz(temp, "a1.dat")
data &lt;- matrix(scan(con),ncol=4,byrow=TRUE)
unlink(temp)
</code></pre>

blocks|key|1006935|text|我使用了在http://cran.r-project.org/web/packages/downloader/index.html找到的CRAN包"downloader“。容易多了。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1006936|download(url,+dest="dataset.zip",+mode="wb")+
unzip+("dataset.zip",+exdir+=+"./")|code-block|syntax|javascript|1006937|entityMap|0|LINK|mutability|MUTABLE|url|http://cran.r-project.org/web/packages/downloader/index.html^0|5|1O|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I used CRAN package "downloader" found at <a href="http://cran.r-project.org/web/packages/downloader/index.html" rel="noreferrer">http://cran.r-project.org/web/packages/downloader/index.html</a> . Much easier.

<pre><code>download(url, dest="dataset.zip", mode="wb") 
unzip ("dataset.zip", exdir = "./")
</code></pre>

blocks|key|1006995|text|对于Mac+(我假设是Linux)...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1006996|如果压缩包只包含一个文件，您可以使用bash命令funzip，与data.table包中的fread结合使用：|offset|length|style|CODE|1006997|library(data.table)
dt+<-+fread("curl+http://www.newcl.org/data/zipfiles/a1.zip+%7C+funzip")|code-block|syntax|javascript|1006998|在归档包含多个文件的情况下，您可以使用tar将特定文件提取到标准输出：|1006999|dt+<-+fread("curl+http://www.newcl.org/data/zipfiles/a1.zip+%7C+tar+-xf-+--to-stdout+*a1.dat")|1007000|entityMap^0|0|O|6|W|A|19|5|0|0|J|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|14|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

For Mac (and I assume Linux)...

If the zip archive contains a single file, you can use the bash command <code>funzip</code>, in conjuction with <code>fread</code> from the <code>data.table</code> package:

<pre><code>library(data.table)
dt &lt;- fread("curl http://www.newcl.org/data/zipfiles/a1.zip | funzip")
</code></pre>

In cases where the archive contains multiple files, you can use <code>tar</code> instead to extract a specific file to stdout:

<pre><code>dt &lt;- fread("curl http://www.newcl.org/data/zipfiles/a1.zip | tar -xf- --to-stdout *a1.dat")
</code></pre>

blocks|key|1221368|text|下面是一个处理无法使用read.table函数读入的文件的示例。此示例读取一个.xls文件。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1221369|url+<-"https://www1.toronto.ca/City_Of_Toronto/Information_Technology/Open_Data/Data_Sets/Assets/Files/fire_stns.zip"

temp+<-+tempfile()
temp2+<-+tempfile()

download.file(url,+temp)
unzip(zipfile+=+temp,+exdir+=+temp2)
data+<-+read_xls(file.path(temp2,+"fire+station+x_y.xls"))

unlink(c(temp,+temp2))|code-block|syntax|javascript|1221370|entityMap^0|B|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Here is an example that works for files which cannot be read in with the <code>read.table</code> function. This example reads a .xls file.

<pre><code>url &lt;-"https://www1.toronto.ca/City_Of_Toronto/Information_Technology/Open_Data/Data_Sets/Assets/Files/fire_stns.zip"

temp &lt;- tempfile()
temp2 &lt;- tempfile()

download.file(url, temp)
unzip(zipfile = temp, exdir = temp2)
data &lt;- read_xls(file.path(temp2, "fire station x_y.xls"))

unlink(c(temp, temp2))
</code></pre>

blocks|key|1221353|text|为了使用data.table实现这一点，我发现下面的方法是可行的。不幸的是，链接不再起作用，所以我使用了另一个数据集的链接。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1221354|library(data.table)
temp+<-+tempfile()
download.file("https://www.bls.gov/tus/special.requests/atusact_0315.zip",+temp)
timeUse+<-+fread(unzip(temp,+files+=+"atusact_0315.dat"))
rm(temp)|code-block|syntax|javascript|1221355|我知道这在一行代码中是可能的，因为您可以将bash脚本传递给fread，但是我不确定如何下载、提取.zip文件，并从中将单个文件传递给fread。|offset|length|style|CODE|1221356|entityMap^0|0|0|U|5|1V|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

To do this using data.table, I found that the following works. Unfortunately, the link does not work anymore, so I used a link for another data set.

<pre><code>library(data.table)
temp &lt;- tempfile()
download.file("https://www.bls.gov/tus/special.requests/atusact_0315.zip", temp)
timeUse &lt;- fread(unzip(temp, files = "atusact_0315.dat"))
rm(temp)
</code></pre>

I know this is possible in a single line since you can pass bash scripts to <code>fread</code>, but I am not sure how to download a .zip file, extract, and pass a single file from that to <code>fread</code>.

blocks|key|1221243|text|试试这段代码。它对我来说很有效：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1221244|unzip(zipfile="<directory+and+filename>",
++++++exdir="<directory+where+the+content+will+be+extracted>")|code-block|syntax|javascript|1221245|示例：|1221246|unzip(zipfile="./data/Data.zip",exdir="./data")|1221247|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Try this code. It works for me:

<pre><code>unzip(zipfile="&lt;directory and filename&gt;",
 exdir="&lt;directory where the content will be extracted&gt;")
</code></pre>

Example:

<pre><code>unzip(zipfile="./data/Data.zip",exdir="./data")
</code></pre>

blocks|key|1221406|text|我发现下面的方法对我很有效。这些步骤来自BTD的YouTube视频Managing+Zipfile's+in+R|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1221407|zip.url+<-+"url_address.zip"

dir+<-+getwd()

zip.file+<-+"file_name.zip"

zip.combine+<-+as.character(paste(dir,+zip.file,+sep+=+"/"))

download.file(zip.url,+destfile+=+zip.combine)

unzip(zip.file)|code-block|syntax|javascript|1221408|entityMap|0|LINK|mutability|MUTABLE|url|https://www.youtube.com/watch?v=fcW8ZFs5sGA^0|X|N|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

I found that the following worked for me. These steps come from BTD's YouTube video, <a href="https://www.youtube.com/watch?v=fcW8ZFs5sGA" rel="nofollow noreferrer">Managing Zipfile's in R</a>:
<pre><code>zip.url &lt;- &quot;url_address.zip&quot;

dir &lt;- getwd()

zip.file &lt;- &quot;file_name.zip&quot;

zip.combine &lt;- as.character(paste(dir, zip.file, sep = &quot;/&quot;))

download.file(zip.url, destfile = zip.combine)

unzip(zip.file)
</code></pre>

blocks|key|1221442|text|rio()非常适合这一点-它使用文件名的文件扩展名来确定它是什么类型的文件，因此它可以处理大量不同的文件类型。我还使用unzip()列出了压缩文件中的文件名，因此没有必要手动指定文件名。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1221443|library(rio)

#+create+a+temporary+directory
td+<-+tempdir()

#+create+a+temporary+file
tf+<-+tempfile(tmpdir=td,+fileext=".zip")

#+download+file+from+internet+into+temporary+location
download.file("http://download.companieshouse.gov.uk/BasicCompanyData-part1.zip",+tf)

#+list+zip+archive
file_names+<-+unzip(tf,+list=TRUE)

#+extract+files+from+zip+file
unzip(tf,+exdir=td,+overwrite=TRUE)

#+use+when+zip+file+has+only+one+file
data+<-+import(file.path(td,+file_names$Name[1]))

#+use+when+zip+file+has+multiple+files
data_multiple+<-+lapply(file_names$Name,+function(x)+import(file.path(td,+x)))

#+delete+the+files+and+directories
unlink(td)|code-block|syntax|javascript|1221444|entityMap^0|0|5|1N|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

<code>rio()</code> would be very suitable for this - it uses the file extension of a file name to determine what kind of file it is, so it will work with a large variety of file types. I've also used <code>unzip()</code> to list the file names within the zip file, so its not necessary to specify the file name(s) manually.
<pre><code>library(rio)

# create a temporary directory
td &lt;- tempdir()

# create a temporary file
tf &lt;- tempfile(tmpdir=td, fileext=&quot;.zip&quot;)

# download file from internet into temporary location
download.file(&quot;http://download.companieshouse.gov.uk/BasicCompanyData-part1.zip&quot;, tf)

# list zip archive
file_names &lt;- unzip(tf, list=TRUE)

# extract files from zip file
unzip(tf, exdir=td, overwrite=TRUE)

# use when zip file has only one file
data &lt;- import(file.path(td, file_names$Name[1]))

# use when zip file has multiple files
data_multiple &lt;- lapply(file_names$Name, function(x) import(file.path(td, x)))

# delete the files and directories
unlink(td)
</code></pre>

@EZGraphs on Twitter writes:
"Lots of online csvs are zipped. Is there a way to download, unzip the archive, and load the data to a data.frame using R? #Rstats"

I was also trying to do this today, but ended up just downloading the zip file manually.

I tried something like:

<pre><code>fileName &lt;- "http://www.newcl.org/data/zipfiles/a1.zip"
con1 &lt;- unz(fileName, filename="a1.dat", open = "r")
</code></pre>

but I feel as if I'm a long way off.
Any thoughts?

Using R to download zipped data file, extract, and import data

Linux

@EZGraphs在推特上写道：“很多在线data.frame都是压缩的。有没有办法下载、解压缩档案，然后使用R将数据加载到csvs？#Rstats”我今天也在尝试这样做，但最终只是手动下载了zip文件。我试过这样的方法：fileName <- "http://www.newcl.org/data/zipfiles/a1.zip"con1 <- unz(fileName, filename="a1

问使用R下载压缩数据文件、提取和导入数据
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