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|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|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|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|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|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

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

@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/a...

问使用R下载压缩数据文件、提取和导入数据
EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问使用R下载压缩数据文件、提取和导入数据EN

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问使用R下载压缩数据文件、提取和导入数据
EN