Your 'x' has length 2, so x[[3]] cannot be calculated ('subscript out of
bounds' is what I get). You can check for this with length(x)3.
In general, you want to be more precise: 'does not have a value', 'is
NULL', and 'is empty' are not synonymous. I'm not sure what 'does not have
a value'
Hello,
Kindly I have a list of lists as follow
x
[[1]]
[1] 7
[[2]]
[1] 3 4 5
as showen x[[3]] does not have a value and it has NULL, how can I check on this
how to test if x[[3]] is empty.
thanks in advance
Ragia
[[alternative HTML version
Hi,
On Dec 20, 2014, at 10:58 AM, Ragia Ibrahim ragi...@hotmail.com wrote:
Hello,
Kindly I have a list of lists as follow
x
[[1]]
[1] 7
[[2]]
[1] 3 4 5
as showen x[[3]] does not have a value and it has NULL, how can I check on
this
how to test if x[[3]] is empty.
In general
Hello,
Your list seems to have only 2 elements. You can check this with
length(x)
Or you can try
lapply(x, is.null)
Hope this helps,
Rui Barradas
Em 20-12-2014 15:58, Ragia Ibrahim escreveu:
Hello,
Kindly I have a list of lists as follow
x
[[1]]
[1] 7
[[2]]
[1] 3 4 5
as showen x[[3]]
This can be tricky, because depending on what the missing object is, you can
get either NULL, NA, or an error. Moreover is.na() behaves differently when
evaluated on its own, or as the condition of an if() statement. Here is a
function that may make life easier. The goal is NOT to have to pass
Boris et. al:
Indeed, corner cases are a bear, which is why it is incumbent on any
OP to precisely define what they mean by, say, missing,
null,empty, etc.
Here is an evil example to illustrate the sorts of nastiness that can occur:
z - list(a=NULL, b=list(), c=NA)
with(z,{
+
This may be out of context, but on the face of it, this claim is wrong:
On 20/12/2014, 1:57 PM, Boris Steipe wrote:
Moreover is.na() behaves differently when evaluated on its own, or as
the condition of an if() statement.
The conditions in an if() statement are not evaluated in special
Thanks. This is what I was referring to:
x - rep(NA, 3)
is.na(x)
[1] TRUE TRUE TRUE
if (is.na(x)) {print(True)}
[1] True
Warning message:
In if (is.na(x)) { :
the condition has length 1 and only the first element will be used
You are of course right - the warning is generated by if(), not by
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Munjal Patel
Sent: Monday, July 14, 2014 8:45 PM
To: r-help@r-project.org
Subject: [R] List of Lists by for Loop
Dear Experts,
I have a one more doubt about making list
Munjal,
Something like this should work:
Digits = c(20, 30, 40, 50, 60)
result = vector(length(Digits), mode=list)
names(result) = Digits
for(j in seq(Digits)) {
a = vector(Digits[j], mode=list)
b = vector(Digits[j], mode=list)
for(i in 1:Digits[j]) {
#Do Calculation
a[[i]] = data.frame()
Dear Experts,
I have a one more doubt about making list of lists.
Here is the simple code i have made.
I am doing the following for only one digit=20
a=vector(20,mode=list)
b=vector(20,mode=list)
for (i in 1:20){
#Do Calculation
a[[i]]=data.frame()
b[[i]]=data.frame()
}
Now i have
Dear Experts,
I have a one more doubt about making list of lists.
Here is the simple code i have made.
I am doing the following for only one digit=20
a=vector(20,mode=list)
b=vector(20,mode=list)
for (i in 1:20){
#Do Calculation
a[[i]]=data.frame()
b[[i]]=data.frame()
}
Now i have to
Re: [R] List of lists
Jim Lemon
of class c
('integer', 'numeric')
Thanks,
Mohan
Re: [R] List
Hi,
I am creating a list of 2 lists, one containing filenames
and the other file descriptors. When I retrieve them I am unable to close
the file descriptor.
I am getting this error when I try to call close(filedescriptors
[[2]][[1]]).
Error in UseMethod(close) :
no applicable
On 07/30/2013 10:05 PM, mohan.radhakrish...@polarisft.com wrote:
Hi,
I am creating a list of 2 lists, one containing filenames
and the other file descriptors. When I retrieve them I am unable to close
the file descriptor.
I am getting this error when I try to call
ON Canada
-Original Message-
From: eliza_bo...@hotmail.com
Sent: Mon, 7 Jan 2013 16:13:03 +
To: r-help@r-project.org
Subject: [R] list of lists to matrix
dear R family,
[a text file has been attached for better understanding]
i have a list of 16 and each of of that is further
dear R family,
[a text file has been attached for better understanding]
i have a list of 16 and each of of that is further subdivided into variable
number of lists. So, i have a kind of list into lists phenomenon.
[[1]]$'1'
1 2 3 4 5 6
7 8 9
[[1]]$'2'
1
Something like
do.call(cbind, lists)
?
MW
On Mon, Jan 7, 2013 at 4:13 PM, eliza botto eliza_bo...@hotmail.com wrote:
dear R family,
[a text file has been attached for better understanding]
i have a list of 16 and each of of that is further subdivided into variable
number of lists. So, i
Thanks arun and weylandt,it perfectly worked out..
elisa
From: michael.weyla...@gmail.com
Date: Mon, 7 Jan 2013 16:37:53 +
Subject: Re: [R] list of lists to matrix
To: eliza_bo...@hotmail.com
CC: r-help@r-project.org
Something like
do.call(cbind, lists)
?
MW
On Mon, Jan 7
: Re: [R] List of lists to data frame?
unlist(..., recursive = F)
Michael
On Wed, Nov 16, 2011 at 6:20 PM, rkevinbur...@charter.net wrote:
I would like to make the following faster:
df - NULL
for(i in 1:length(s))
{
df - rbind(df, cbind(names(s[i
I would like to make the following faster:
df - NULL
for(i in 1:length(s))
{
df - rbind(df, cbind(names(s[i]), time(s[[i]]$series),
as.vector(s[[i]]$series), s[[i]]$category))
}
names(df) - c(name, time, value, category)
return(df)
unlist(..., recursive = F)
Michael
On Wed, Nov 16, 2011 at 6:20 PM, rkevinbur...@charter.net wrote:
I would like to make the following faster:
df - NULL
for(i in 1:length(s))
{
df - rbind(df, cbind(names(s[i]), time(s[[i]]$series),
to the master data
frame but like I said it is *very* slow.
Kevin
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Wednesday, November 16, 2011 5:26 PM
To: rkevinbur...@charter.net
Cc: r-help@r-project.org
Subject: Re: [R] List of lists to data frame?
unlist
, 2011 5:26 PM
To: rkevinbur...@charter.net
Cc: r-help@r-project.org
Subject: Re: [R] List of lists to data frame?
unlist(..., recursive = F)
Michael
On Wed, Nov 16, 2011 at 6:20 PM, rkevinbur...@charter.net wrote:
I would like to make the following faster:
df - NULL
On the face of it this looks like a job for ldply() in the plyr package
which specialises in taking things apart and putting them back together.
ldply() applies a function for each element of a list and then combine
results into a data frame
On 17 November 2011 04:53, Sarah Goslee
Dear list,
I have to use a list of lists containing vectors.
For instance :
[[1]]
[[1]][[1]]
[1] 1 2 3
[[1]][[2]]
[1] 3 2 1
I want to attribute vectors to the main list
without use of an intermediate list,
but it does not work :
x - list()
x[[1]][[1]] - c(1, 2, 3)
x[[1]][[2]] - c(3, 2, 1)
On Aug 9, 2010, at 12:57 PM, Carlos Petti wrote:
Dear list,
I have to use a list of lists containing vectors.
For instance :
[[1]]
[[1]][[1]]
[1] 1 2 3
[[1]][[2]]
[1] 3 2 1
I want to attribute vectors to the main list
without use of an intermediate list,
but it does not work :
More
Is this what you want:
x - list()
x[[1]] - list(1:3)
x[[2]] - list(3:1)
x
[[1]]
[[1]][[1]]
[1] 1 2 3
[[2]]
[[2]][[1]]
[1] 3 2 1
On Mon, Aug 9, 2010 at 12:57 PM, Carlos Petti carlos.pe...@gmail.com wrote:
Dear list,
I have to use a list of lists containing vectors.
For instance :
See if one of %in% or match gets your further.
1:10 %in% c(1,3,5,9)
[1] TRUE FALSE TRUE FALSE TRUE FALSE FALSE FALSE TRUE FALSE
match(c(1,3,5,9), 1:10)
[1] 1 3 5 9
match(c(1,3,5,9), 10:1)
[1] 10 8 6 2
date03 as offered was not a list, but a vector.
date04 - date02[which(date02$date
On Wed, Jan 14, 2009 at 1:53 PM, glenn g1enn.robe...@btinternet.com wrote:
Dear All;
Is it possible to create a list of lists (I am sure it is) along these
lines;
I have a dataframe data02 that holds a lot of information, and the first
column is ³date²
I have a list of dates in;
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