string 从R中的字符串中提取唯一数字
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Extracting unique numbers from string in R
提问by Remi.b
I have a list of strings which contain random characters such as:
我有一个包含随机字符的字符串列表，例如：
list=list()
list[1] = "djud7+dg[a]hs667"
list[2] = "7fd*hac11(5)"
list[3] = "2tu,g7gka5"
I'd like to know which numbers are present at least once (unique()) in this list. The solution of my example is:
我想知道unique()在此列表中哪些数字至少出现了一次 ( )。我的例子的解决方案是：
solution: c(7,667,11,5,2)
解决方案： c(7,667,11,5,2)
If someone has a method that does not consider 11 as "eleven" but as "one and one", it would also be useful. The solution in this condition would be:
如果有人有一种不将 11 视为“十一”而是“一加一”的方法，它也会很有用。在这种情况下的解决方案是：
solution: c(7,6,1,5,2)
解决方案： c(7,6,1,5,2)
(I found this post on a related subject: Extracting numbers from vectors of strings)
（我在相关主题上找到了这篇文章：从字符串向量中提取数字）
回答by Arun
For the second answer, you can use gsubto remove everything from the string that's not a number, then split the string as follows:
对于第二个答案，您可以使用gsub从字符串中删除不是数字的所有内容，然后按如下方式拆分字符串：
unique(as.numeric(unlist(strsplit(gsub("[^0-9]", "", unlist(ll)), ""))))
# [1] 7 6 1 5 2
For the first answer, similarly using strsplit,
对于第一个答案，同样使用strsplit，
unique(na.omit(as.numeric(unlist(strsplit(unlist(ll), "[^0-9]+")))))
# [1] 7 667 11 5 2
PS: don't name your variable list(as there's an inbuilt function list). I've named your data as ll.
PS：不要命名你的变量list（因为有一个内置函数list）。我已将您的数据命名为ll.
回答by A5C1D2H2I1M1N2O1R2T1
Here is yet another answer, this one using gregexprto find the numbers, and regmatchesto extract them:
这是另一个答案，这个答案gregexpr用于查找数字并regmatches提取它们：
l <- c("djud7+dg[a]hs667", "7fd*hac11(5)", "2tu,g7gka5")
temp1 <- gregexpr("[0-9]", l) # Individual digits
temp2 <- gregexpr("[0-9]+", l) # Numbers with any number of digits
as.numeric(unique(unlist(regmatches(l, temp1))))
# [1] 7 6 1 5 2
as.numeric(unique(unlist(regmatches(l, temp2))))
# [1] 7 667 11 5 2
回答by altabq
A solution using stringi
使用stringi的解决方案
# extract the numbers:
nums <- stri_extract_all_regex(list, "[0-9]+")
# Make vector and get unique numbers:
nums <- unlist(nums)
nums <- unique(nums)
And that's your first solution
这是你的第一个解决方案
For the second solution I would use substr:
对于第二个解决方案，我将使用substr：
nums_first <- sapply(nums, function(x) unique(substr(x,1,1)))
回答by sgibb
You could use ?strsplit(like suggested in @Arun's answer in Extracting numbers from vectors (of strings)):
您可以使用?strsplit（就像@Arun 在从向量中提取数字（字符串）中的答案中所建议的那样）：
l <- c("djud7+dg[a]hs667", "7fd*hac11(5)", "2tu,g7gka5")
## split string at non-digits
s <- strsplit(l, "[^[:digit:]]")
## convert strings to numeric ("" become NA)
solution <- as.numeric(unlist(s))
## remove NA and duplicates
solution <- unique(solution[!is.na(solution)])
# [1] 7 667 11 5 2
回答by Joe
A stringrsolution with str_match_alland piped operators. For the first solution:
阿stringr与溶液str_match_all和管道运营商。对于第一个解决方案：
library(stringr)
str_match_all(ll, "[0-9]+") %>% unlist %>% unique %>% as.numeric
Second solution:
第二种解决方案：
str_match_all(ll, "[0-9]") %>% unlist %>% unique %>% as.numeric
(Note: I've also called the list ll)
（注意：我也调用了列表ll）
回答by asb
Use strsplit using pattern as the inverse of numeric digits: 0-9
使用 strsplit 使用模式作为数字的倒数：0-9
For the example you have provided, do this:
对于您提供的示例，请执行以下操作：
tmp <- sapply(list, function (k) strsplit(k, "[^0-9]"))
Then simply take a union of all `sets' in the list, like so:
然后简单地取列表中所有“集合”的并集，如下所示：
tmp <- Reduce(union, tmp)
Then you only have to remove the empty string.
然后你只需要删除空字符串。
回答by Rory Nolan
Check out the str_extract_numbers()function from the strexpackage.
查看包中的str_extract_numbers()功能strex。
pacman::p_load(strex)
list=list()
list[1] = "djud7+dg[a]hs667"
list[2] = "7fd*hac11(5)"
list[3] = "2tu,g7gka5"
charvec <- unlist(list)
print(charvec)
#> [1] "djud7+dg[a]hs667" "7fd*hac11(5)" "2tu,g7gka5"
str_extract_numbers(charvec)
#> [[1]]
#> [1] 7 667
#>
#> [[2]]
#> [1] 7 11 5
#>
#> [[3]]
#> [1] 2 7 5
unique(unlist(str_extract_numbers(charvec)))
#> [1] 7 667 11 5 2
Created on 2018-09-03 by the reprex package(v0.2.0).
由reprex 包(v0.2.0)于 2018 年 9 月 3 日创建。
