Dear all,
I have a vector with a certain range of values including infinity and
NA. I would like to remove the values that are outside a given range
(lower level = ll and upper level = ul) but I am getting the error due
to the NA values (missing value where TRUE/FALSE needed). I then
included the !is.na() but now the resulting error is all NA, as in the
example.
In addition, here I have implemented a for loop to scan all the
elements of the vector, but I should be able to use the sapply();
however I don't know how to send the ll and ul arguments to sapply().
Could you please help?
best regards
Luigi
EXAMPLE
x <- c(-Inf, Inf, NA, 5.9, 6.08, 5281391136138.75,
4.35, 4.79,
9474097322.96, 3.64, 16.42, -12211.11, 4.37,
-1097.79, 4.78,
3.71, 32.59, 4.01, 35.36, 3.17, 1.61,
-3678.28, 2.9, 4.67,
4.1, 348410866.78, 5.35, 4.3101519459837E+016,
1467030866.75,
1.10376094956278E+018, 32.55, 1.17, 5339028670388.94,
34.14,
33205967009.57, 4.42, 1.76, 7.08, -8428.84,
-113491.08, 17.81)
ll <- 1
ul <- 45
clipper <- function(x, ll, ul) {
for(i in 1:length(x)) {
if(x[i] < ll) {
x[i] <- NA
} else if(x[i] > ul) {
x[i] <- NA
} else {
x[i] <- x[i]
}
}
return(x)
}
(X<-clipper(x, ll, ul))> missing value where TRUE/FALSE needed
clipper <- function(x, ll, ul) {
for(i in 1:length(x)) {
if(!is.na(x[i]) < ll) {
x[i] <- NA
} else if(!is.na(x[i]) > ul) {
x[i] <- NA
} else {
x[i] <- x[i]
}
}
return(x)
}
(X<-clipper(x, ll, ul))
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
NA NA NA NA NA
[28] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
On Apr 17, 2015, at 5:23 PM, Luigi Marongiu <marongiu.luigi at gmail.com> wrote:> > Dear all, > I have a vector with a certain range of values including infinity and > NA. I would like to remove the values that are outside a given range > (lower level = ll and upper level = ul) but I am getting the error due > to the NA values (missing value where TRUE/FALSE needed). I then > included the !is.na() but now the resulting error is all NA, as in the > example. > In addition, here I have implemented a for loop to scan all the > elements of the vector, but I should be able to use the sapply(); > however I don't know how to send the ll and ul arguments to sapply(). > Could you please help? > best regards > Luigi > > EXAMPLE > > x <- c(-Inf, Inf, NA, 5.9, 6.08, 5281391136138.75, > 4.35, 4.79, > 9474097322.96, 3.64, 16.42, -12211.11, 4.37, > -1097.79, 4.78, > 3.71, 32.59, 4.01, 35.36, 3.17, 1.61, > -3678.28, 2.9, 4.67, > 4.1, 348410866.78, 5.35, 4.3101519459837E+016, > 1467030866.75, > 1.10376094956278E+018, 32.55, 1.17, 5339028670388.94, > 34.14, > 33205967009.57, 4.42, 1.76, 7.08, -8428.84, > -113491.08, 17.81) > ll <- 1 > ul <- 45 > > clipper <- function(x, ll, ul) { > for(i in 1:length(x)) { > if(x[i] < ll) { > x[i] <- NA > } else if(x[i] > ul) { > x[i] <- NA > } else { > x[i] <- x[i] > } > } > return(x) > } > (X<-clipper(x, ll, ul)) >> missing value where TRUE/FALSE needed > > > clipper <- function(x, ll, ul) { > for(i in 1:length(x)) { > if(!is.na(x[i]) < ll) { > x[i] <- NA > } else if(!is.na(x[i]) > ul) { > x[i] <- NA > } else { > x[i] <- x[i] > } > } > return(x) > } > (X<-clipper(x, ll, ul)) > > [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA > NA NA NA NA NA > [28] NA NA NA NA NA NA NA NA NA NA NA NA NA NAHi, Something along the lines of:> subset(x, is.finite(x) & (x > ll) & (x < ul))[1] 5.90 6.08 4.35 4.79 3.64 16.42 4.37 4.78 3.71 32.59 4.01 [12] 35.36 3.17 1.61 2.90 4.67 4.10 5.35 32.55 1.17 34.14 4.42 [23] 1.76 7.08 17.81 or:> x[is.finite(x) & (x > ll) & (x < ul)][1] 5.90 6.08 4.35 4.79 3.64 16.42 4.37 4.78 3.71 32.59 4.01 [12] 35.36 3.17 1.61 2.90 4.67 4.10 5.35 32.55 1.17 34.14 4.42 [23] 1.76 7.08 17.81 See ?subset and ?is.finite:> is.finite(x)[1] FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [34] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE Regards, Marc Schwartz
Dear David and Mark,
thank you for your reply. I have implemented the suggestions you have
made in the following:
x <- c(-Inf, Inf, NA, 5.9, 6.08, 5281391136138.75,
4.35, 4.79,
9474097322.96, 3.64, 16.42, -12211.11, 4.37,
-1097.79, 4.78,
3.71, 32.59, 4.01, 35.36, 3.17, 1.61,
-3678.28, 2.9, 4.67,
4.1, 348410866.78, 5.35, 4.3101519459837E+016,
1467030866.75,
1.10376094956278E+018, 32.55, 1.17, 5339028670388.94,
34.14,
33205967009.57, 4.42, 1.76, 7.08, -8428.84,
-113491.08, 17.81)
ll <- 1
ul <- 45
clipper <- function(x, ll, ul) {
for(i in 1:length(x)) {
if(is.finite(x[i]) < ll & is.finite(x[i]) > ul) {
x[i] <- NA
} else if (is.infinite(x[i]) == "TRUE") {
x[i] <- NA
} else {
x[i] <- x[i]
}
}
return(x)
}
(X<-clipper(x, ll, ul))
that works all right.
Best regards
Luigi
On Fri, Apr 17, 2015 at 11:43 PM, Marc Schwartz <marc_schwartz at me.com>
wrote:> On Apr 17, 2015, at 5:23 PM, Luigi Marongiu <marongiu.luigi at
gmail.com> wrote:
>>
>> Dear all,
>> I have a vector with a certain range of values including infinity and
>> NA. I would like to remove the values that are outside a given range
>> (lower level = ll and upper level = ul) but I am getting the error due
>> to the NA values (missing value where TRUE/FALSE needed). I then
>> included the !is.na() but now the resulting error is all NA, as in the
>> example.
>> In addition, here I have implemented a for loop to scan all the
>> elements of the vector, but I should be able to use the sapply();
>> however I don't know how to send the ll and ul arguments to
sapply().
>> Could you please help?
>> best regards
>> Luigi
>>
>> EXAMPLE
>>
>> x <- c(-Inf, Inf, NA, 5.9, 6.08, 5281391136138.75,
>> 4.35, 4.79,
>> 9474097322.96, 3.64, 16.42, -12211.11, 4.37,
>> -1097.79, 4.78,
>> 3.71, 32.59, 4.01, 35.36, 3.17, 1.61,
>> -3678.28, 2.9, 4.67,
>> 4.1, 348410866.78, 5.35, 4.3101519459837E+016,
>> 1467030866.75,
>> 1.10376094956278E+018, 32.55, 1.17, 5339028670388.94,
>> 34.14,
>> 33205967009.57, 4.42, 1.76, 7.08, -8428.84,
>> -113491.08, 17.81)
>> ll <- 1
>> ul <- 45
>>
>> clipper <- function(x, ll, ul) {
>> for(i in 1:length(x)) {
>> if(x[i] < ll) {
>> x[i] <- NA
>> } else if(x[i] > ul) {
>> x[i] <- NA
>> } else {
>> x[i] <- x[i]
>> }
>> }
>> return(x)
>> }
>> (X<-clipper(x, ll, ul))
>>> missing value where TRUE/FALSE needed
>>
>>
>> clipper <- function(x, ll, ul) {
>> for(i in 1:length(x)) {
>> if(!is.na(x[i]) < ll) {
>> x[i] <- NA
>> } else if(!is.na(x[i]) > ul) {
>> x[i] <- NA
>> } else {
>> x[i] <- x[i]
>> }
>> }
>> return(x)
>> }
>> (X<-clipper(x, ll, ul))
>>
>> [1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
>> NA NA NA NA NA
>> [28] NA NA NA NA NA NA NA NA NA NA NA NA NA NA
>
>
> Hi,
>
> Something along the lines of:
>
>> subset(x, is.finite(x) & (x > ll) & (x < ul))
> [1] 5.90 6.08 4.35 4.79 3.64 16.42 4.37 4.78 3.71 32.59 4.01
> [12] 35.36 3.17 1.61 2.90 4.67 4.10 5.35 32.55 1.17 34.14 4.42
> [23] 1.76 7.08 17.81
>
> or:
>
>> x[is.finite(x) & (x > ll) & (x < ul)]
> [1] 5.90 6.08 4.35 4.79 3.64 16.42 4.37 4.78 3.71 32.59 4.01
> [12] 35.36 3.17 1.61 2.90 4.67 4.10 5.35 32.55 1.17 34.14 4.42
> [23] 1.76 7.08 17.81
>
>
> See ?subset and ?is.finite:
>
>> is.finite(x)
> [1] FALSE FALSE FALSE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> [12] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> [23] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
> [34] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>
>
> Regards,
>
> Marc Schwartz
>