[英]How to write multiple if statement R
I am manipulating raster data in R with the rgdal
and raster
packages.我正在使用rgdal
和raster
包在 R 中操作光栅数据。 I want to get rid of all infinite, no values, negative values and replace them with zero:我想摆脱所有无限、无值、负值并用零替换它们:
NoNA <- function (x) {
x[is.infinite(x) | is.na(x) | is.nan(x) | is.null(x) | is.na(x < 0)] <- 0
}
ndii_noNA <- calc(ndii, NoNA)
Then the ndii_noNA
have only a value of 0. I tried if else statement but it raises an error in然后ndii_noNA
的值只有 0。我试过 if else 语句,但它引发了一个错误
.calcTest(x[1:5], fun, na.rm, forcefun, forceapply).
Is there any way to solve this?有没有办法解决这个问题?
You are very close, but have made two errors:你非常接近,但犯了两个错误:
which()
in the index of x
, rather than just the truth statement.您需要在x
的索引中使用which()
,而不仅仅是真相陈述。 Otherwise, you will index either x[TRUE]
or x[FALSE]
, which is not what you want.否则,您将索引x[TRUE]
或x[FALSE]
,这不是您想要的。 which()
will return the indices of all "bad" elements in your vector. which()
将返回向量中所有“坏”元素的索引。<-
, the local copy of x
will be changed, not the one that was passed.当您使用<-
进行分配时, x
的本地副本将被更改,而不是通过的副本。 If you want to change x
in place, you need to use <<-
.如果要原地更改x
,则需要使用<<-
。 That said, you would be wise to stick to R's functional paradigm, in which you would make the change to the local copy, then return it with return(x)
, rather than change in place.也就是说,您最好坚持 R 的功能范式,在这种范式中,您将对本地副本进行更改,然后使用return(x)
返回它,而不是原地更改。Here is the function you want:这是你想要的功能:
# The "change in place" method (may be considered bad style)
NoNA <- function(x) {
x[which(is.infinite(x) | is.na(x) | is.nan(x) | is.null(x) | is.na(x < 0))] <<- 0
}
# The functional way (recommended)
NoNA <- function(x) {
x[which(is.infinite(x) | is.na(x) | is.nan(x) | is.null(x) | is.na(x < 0))] <- 0
return(x)
}
Edit: ifelse() is cleaner but the @cgmil's answer is indeed faster.编辑: ifelse() 更干净,但@cgmil 的答案确实更快。
x = rep(c(Inf, -Inf, NULL, NaN, NA, 1), 250e3)
no_na = function(x){
ifelse(
is.infinite(x) | is.na(x) | is.nan(x) | is.null(x) | is.na(x < 0), 0, x
)
}
NoNA <- function(x) {
x[which(is.infinite(x) | is.na(x) | is.nan(x) | is.null(x) | is.na(x < 0))] <- 0
return(x)
}
microbenchmark(
no_na(x), NoNA(x),
times = 50
)
# Unit: milliseconds
# expr min lq mean median uq max neval
# no_na(x) 380.9375 399.7520 416.7729 424.5490 429.6005 451.0534 50
# NoNA(x) 242.8555 249.0034 255.8857 251.3694 254.8176 285.1451 50
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