问题
I'd like to grep for "nitrogen" in the following character vector and want to get back only the entry which is containing "nitrogen" and nothing of the rest (e.g. nitrogen fixation):
varnames=c("nitrogen", "dissolved organic nitrogen", "nitrogen fixation", "total dissolved nitrogen", "total nitrogen")
I tried something like this:
grepl(pattern= "![[:space:]]nitrogen![[:space:]]", varnames)
But this doesn't work. I hope somebody can help me with this task.
回答1:
To get the indices that are exactly equal to "nitrogen" you could use
which(varnames == "nitrogen")
Depending on what you want to do you might not even need the 'which' as varnames == "nitrogen"
gives a logical vector of TRUE/FALSE. If you just want to do something like replace all of the occurances of "nitrogen" with "oxygen" this should suffice
varnames[varnames == "nitrogen"] <- "oxygen"
回答2:
Although Dason's answer is easier, you could do an exact match using grep via:
varnames=c("nitrogen", "dissolved organic nitrogen", "nitrogen fixation", "total dissolved nitrogen", "total nitrogen")
grep("^nitrogen$",varnames,value=TRUE)
[1] "nitrogen"
grep("^nitrogen$",varnames)
[1] 1
回答3:
Or use fixed = TRUE
if you want to match actual string (regex
lessly):
v <- sample(c("nitrogen", "potassium", "hidrogen"), size = 100, replace = TRUE, prob = c(.8, .1, .1))
grep("nitrogen", v, fixed = TRUE)
# [1] 3 4 5 6 7 8 9 11 12 13 14 16 19 20 21 22 23 24 25
# [20] 26 27 29 31 32 35 36 38 39 40 41 43 44 46 47 48 49 50 51
# [39] 52 53 54 56 57 60 61 62 65 66 67 69 70 71 72 73 74 75 76
# [58] 78 79 80 81 82 83 84 85 86 87 88 89 91 92 93 94 95 96 97
# [77] 98 99 100
Dunno about the speed issues, I like to test stuff and claim that approach A is faster than approach B, but in theory, at least from my experience, indexing/binary operators should be the fastest, so I vote for @Dason's approach. Also note that regex
es are always slower than fixed = TRUE
grep
ing.
A little proof is attached bellow. Note that this is a lame test, and system.time
should be put inside replicate
to get (more) accurate differences, you should take outliers into an account, etc. But surely this one proves that you should use which
! =)
(a0 <- system.time(replicate(1e5, grep("^nitrogen$", v))))
# user system elapsed
# 5.700 0.023 5.724
(a1 <- system.time(replicate(1e5, grep("nitrogen", v, fixed = TRUE))))
# user system elapsed
# 1.147 0.020 1.168
(a2 <- system.time(replicate(1e5, which(v == "nitrogen"))))
# user system elapsed
# 1.013 0.020 1.033
来源:https://stackoverflow.com/questions/10041810/how-to-grep-a-word-exactly