[R] if else statement for rain data to define zero for dry and one to wet
roslinazairimah zakaria
roslinaump at gmail.com
Sun Jun 7 16:57:52 CEST 2015
Dear all,
All works well. Thank you so much for your help.
D## Function 1
wet_dry1 <- function(x,thresh=0.1)
{ for(column in 1:dim(x)[2]) x[,column] <- ifelse(x[,column]>=thresh,1,0)
return(x)
}
wet_dry1(dt)
## Function 2
wet_dry2 <- ( dt >= 0.1)*1
wet_dry2
wet_total <- colSums(wet_dry2)
pp <- wet_total/nrow(dt)
pp
## Function 3
rain <- dt
wet_dry3 <- ifelse(rain >= 0.1, 1, 0)
wet_dry3
On Sun, Jun 7, 2015 at 5:48 AM, William Dunlap <wdunlap at tibco.com> wrote:
> Your f1() has an unneeded for loop in it.
> f1a <- function(mat) mat > 0.1, 1, 0)
> would do the same thing in a bit less time.
>
> However, I think that a simple
> mat > 0.1
> would be preferable. The resulting TRUEs and FALSEs
> are easier to interpret than the 1s and 0s that f1a()
> produces and arithmetic functions treat them TRUE
> as 1 and FALSE as 0 internally. E.g., mean(mat>0.1)
> gives the proportion of wet(tish) days.
>
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Sat, Jun 6, 2015 at 1:55 PM, Dennis Murphy <djmuser at gmail.com> wrote:
>
>> I'm sorry, but I have to take issue with this particular use case of
>> ifelse(). When the goal is to generate a logical vector, ifelse() is
>> very inefficient. It's better to apply a logical condition directly to
>> the object in question and multiply the result by 1 to make it
>> numeric/integer rather than logical.
>>
>> To illustrate this, consider the following toy example. The function
>> f1 replicates the suggestion to apply ifelse() columnwise (with the
>> additional overhead of preallocating storage for the result), whereas
>> the function f2 applies the logical condition on the matrix itself
>> using vectorization, with the recognition that a matrix is an atomic
>> vector with a dim attribute.
>>
>> set.seed(5290)
>>
>> # 1000 x 1000 matrix
>> m <- matrix(sample(c(0, 0.05, 0.2), 1e6, replace = TRUE), ncol = 1000)
>>
>> f1 <- function(mat)
>> {
>> newmat <- matrix(NA, ncol = ncol(mat), nrow = nrow(mat))
>> for(i in seq_len(ncol(mat)))
>> newmat[, i] <- ifelse(mat[, i] > 0.1, 1, 0)
>> newmat
>> }
>>
>> f2 <- function(mat) 1 * (mat > 0.1)
>>
>>
>> On my system, I got
>>
>> > system.time(m1 <- f1(m))
>> user system elapsed
>> 0.14 0.00 0.14
>>
>> > system.time(m2 <- f2(m))
>> user system elapsed
>> 0.01 0.00 0.01
>>
>> > identical(m1, m2)
>> [1] TRUE
>>
>> The all too common practice of using ifelse(condition, 1, 0) on an
>> atomic vector is easily replaced by 1 * (condition), where the result
>> of condition is a logical atomic object coerced to numeric.
>>
>> To reduce memory, one should better define f2 as
>>
>> f2 <- function(mat) 1L * (mat > 0.1)
>>
>> but doing so in this example no longer creates identical objects since
>>
>> > typeof(m1)
>> [1] "double"
>>
>> Thus, f1 is not only inefficient in terms of execution time, it's also
>> inefficient in terms of storage.
>>
>> Given several recent warnings in this forum about the inefficiency of
>> ifelse() and the dozens of times I've seen the idiom implemented in f1
>> as a solution over the last several years (to which I have likely
>> contributed in my distant past as an R-helper), I felt compelled to
>> say something about this practice, which BTW extends not just to 0/1
>> return values but to
>> 0/x return values, where x is a nonzero real number.
>>
>> Dennis
>>
>>
>> On Sat, Jun 6, 2015 at 12:50 AM, Jim Lemon <drjimlemon at gmail.com> wrote:
>> > Hi rosalinazairimah,
>> > I think the problem is that you are using "if" instead of "ifelse". Try
>> this:
>> >
>> > wet_dry<-function(x,thresh=0.1) {
>> > for(column in 1:dim(x)[2]) x[,column]<-ifelse(x[,column]>=thresh,1,0)
>> > return(x)
>> > }
>> > wet_dry(dt)
>> >
>> > and see what you get.
>> >
>> > Also, why can I read your message perfectly while everybody else can't?
>> >
>> > Jim
>> >
>> >>> -----Original Message-----
>> >>> From: roslinaump at gmail.com
>> >>> Sent: Fri, 5 Jun 2015 16:49:08 +0800
>> >>> To: r-help at r-project.org
>> >>> Subject: [R] if else statement for rain data to define zero for dry
>> and
>> >>> one to wet
>> >>>
>> >>> Dear r-users,
>> >>>
>> >>> I have a set of rain data:
>> >>>
>> >>> X1950 X1951 X1952 X1953 X1954 X1955 X1956 X1957 X1958 X1959 X1960
>> X1961
>> >>> X1962
>> >>>
>> >>> 1 0.0 0.0 14.3 0.0 13.5 13.2 4.0 0 3.3 0 0
>> 0.0
>> >>>
>> >>>
>> >>> 2 0.0 0.0 21.9 0.0 10.9 6.6 2.1 0 0.0 0 0
>> 0.0
>> >>>
>> >>>
>> >>> 3 25.3 6.7 18.6 0.8 2.3 0.0 8.0 0 0.0 0 0
>> 11.0
>> >>>
>> >>>
>> >>> 4 12.7 3.4 37.2 0.9 8.4 0.0 5.8 0 0.0 0 0
>> 5.5
>> >>>
>> >>>
>> >>> 5 0.0 0.0 58.3 3.6 21.1 4.2 3.0 0 0.0 0 0
>> 15.9
>> >>>
>> >>>
>> >>> I would like to go through each column and define each cell with value
>> >>> greater than 0.1 mm will be 1 and else zero. Hence I would like to
>> attach
>> >>> the rain data and the category side by side:
>> >>>
>> >>>
>> >>> 1950 state
>> >>>
>> >>> 1 0.0 0
>> >>>
>> >>> 2 0.0 0
>> >>>
>> >>> 3 25.3 1
>> >>>
>> >>> 4 12.7 1
>> >>>
>> >>> 5 0.0 0
>> >>>
>> >>>
>> >>> ...
>> >>>
>> >>>
>> >>> This is my code:
>> >>>
>> >>>
>> >>> wet_dry <- function(dt)
>> >>>
>> >>> { cl <- length(dt)
>> >>>
>> >>> tresh <- 0.1
>> >>>
>> >>>
>> >>> for (i in 1:cl)
>> >>>
>> >>> { xi <- dt[,i]
>> >>>
>> >>> if (xi < tresh ) 0 else 1
>> >>>
>> >>> }
>> >>>
>> >>> dd <- cbind(dt,xi)
>> >>>
>> >>> dd
>> >>>
>> >>> }
>> >>>
>> >>>
>> >>> wet_dry(dt)
>> >>>
>> >>>
>> >>> Results:
>> >>>
>> >>>> wet_dry(dt)
>> >>>
>> >>> X1950 X1951 X1952 X1953 X1954 X1955 X1956 X1957 X1958 X1959 X1960
>> >>> X1961
>> >>> X1962 X1963 X1964 X1965 X1966 X1967 X1968 X1969 X1970 X1971 X1972
>> X1973
>> >>> X1974 X1975 X1976 X1977
>> >>>
>> >>> 1 0.0 0.0 14.3 0.0 13.5 13.2 4.0 0.0 3.3 0.0 0.0
>> >>> 0.0
>> >>> 4.2 0.0 2.2 0.0 4.4 5.1 0 7.2 0.0 0.0 0.0
>> 5.1
>> >>> 0 0.0 0 0.3
>> >>>
>> >>> 2 0.0 0.0 21.9 0.0 10.9 6.6 2.1 0.0 0.0 0.0 0.0
>> >>> 0.0
>> >>> 8.4 0.0 4.0 0.0 4.9 0.7 0 0.0 0.0 0.0 0.0
>> 5.4
>> >>> 0 3.3 0 0.3
>> >>>
>> >>> 3 25.3 6.7 18.6 0.8 2.3 0.0 8.0 0.0 0.0 0.0 0.0
>> >>> 11.0
>> >>> 4.2 0.0 2.0 0.0 14.2 17.1 0 0.0 0.0 0.0 0.0
>> 2.1
>> >>> 0 1.7 0 4.4
>> >>>
>> >>> 4 12.7 3.4 37.2 0.9 8.4 0.0 5.8 0.0 0.0 0.0 0.0
>> >>> 5.5
>> >>> 0.0 0.0 5.4 0.0 6.4 14.9 0 10.1 2.9 143.4 0.0
>> 6.1
>> >>> 0 0.0 0 33.5
>> >>>
>> >>>
>> >>> It does not work and give me the original data. Why is that?
>> >>>
>> >>>
>> >>> Thank you so much for your help.
>> >>>
>> >>> [[alternative HTML version deleted]]
>> >>>
>> >>> ______________________________________________
>> >>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> >>> https://stat.ethz.ch/mailman/listinfo/r-help
>> >>> PLEASE do read the posting guide
>> >>> http://www.R-project.org/posting-guide.html
>> >>> and provide commented, minimal, self-contained, reproducible code.
>> >
>> > ______________________________________________
>> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
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