Rei*_*son 14
比刮取网页以获取包名称更好的方法是使用该available.packages()功能并处理这些结果.available.packages()返回一个矩阵,其中包含所有可用包的详细信息(但默认情况下会过滤 - 有关详细信息,请参阅详细信息部分?available.packages).
pkgs <- available.packages(filters = "duplicates")
nameCount <- unname(nchar(pkgs[, "Package"]))
table(nameCount)
> table(nameCount)
nameCount
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
32 311 374 360 434 445 368 277 199 132 99 56 56 43 22 19 18 2 12 8
22 24 25 31
5 2 1 1
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使用nameCount我们可以选择包含任意数量字符的包,而无需使用regexp等:
> unname(pkgs[which(nameCount == 2), "Package"])
[1] "BB" "bs" "ca" "cg" "dr" "ez" "FD" "ff" "HH" "HI" "iv" "JM" "ks" "M3" "mi"
[16] "np" "oc" "oz" "PK" "PP" "qp" "QT" "RC" "rv" "Rz" "sm" "sn" "sp" "st" "SV"
[31] "tm" "wq"
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bap*_*ste 10
这是基于各种建议的一次拍摄.
packages <- available.packages()[,'Package']
ggplot(data.frame(n = nchar(packages))) +
geom_histogram(aes(n), binwidth=1)
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all <- length(packages)
## 3168
up <- sum(toupper(packages) == packages)
## 262
low <- sum(tolower(packages) == packages)
## 1697
pie(c(up, low, all-up-low), labels=c("UPPERCASE","lowercase","cAmElCaSe"))
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let <- sapply(sapply(letters, grep, tolower(packages)), length)
barplot(let)
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length(packages[grep("2$", packages, perl=TRUE)])
# 29
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这里有一小段代码可以回答一些问题.我找时间的时候会继续补充答案.
library(XML); library(ggplot2);
url = 'http://cran.r-project.org/web/packages/available_packages_by_name.html'
packages = readHTMLTable(url, stringsAsFactors = F)[[1]][-1,]
# histogram of number of characters in package name
qplot(nchar(V1), data = packages)
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