Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have not only 2 vectors, but also 4.
Question is, if q1 and q2 is equal to w1 and w2.
The computational time is very short, also for large data.
q1 - c(9,5,1,5)
q2 - c(9,2,1,5)
w1 - c(9,4,4,4,5)
w1 -
Hey Chris,
I would take advantage from the apply function:
apply(cbind(q1,q2),1,function(x)any((x[1]==w1)(x[2]==w2)))
Regards
PF
On Thu, Feb 2, 2012 at 12:55 PM, Chris82 rubenba...@gmx.de wrote:
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have
On Feb 2, 2012, at 6:55 AM, Chris82 wrote:
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have not only 2 vectors, but also 4.
Question is, if q1 and q2 is equal to w1 and w2.
The computational time is very short, also for large data.
q1 -
Hi
Thanks to Berend and the others,
I've found a solution which works fine for my problem.
I have not only 2 vectors, but also 4.
Question is, if q1 and q2 is equal to w1 and w2.
The computational time is very short, also for large data.
q1 - c(9,5,1,5)
q2 - c(9,2,1,5)
w1 -
Hi R users,
is there any possibilty that a while loop is working like that:
z - c(0,1,2,3,4,5,6,7,8,9)
r - 7
while(w == T) {
for ( i in 1:10 ){
w - r == z[i]
print(w)
}
}
The loop should stop if w == TRUE
best regards
--
View this
You are combining too many loop constructs: perhaps you just want to
use for and break.
Of course, in your case it's much faster to write which(r == z) or
which.min(r == z)
Michael
On Wed, Feb 1, 2012 at 10:55 AM, Chris82 rubenba...@gmx.de wrote:
Hi R users,
is there any possibilty that a
Le mercredi 01 février 2012 à 07:55 -0800, Chris82 a écrit :
Hi R users,
is there any possibilty that a while loop is working like that:
z - c(0,1,2,3,4,5,6,7,8,9)
r - 7
while(w == T) {
for ( i in 1:10 ){
w - r == z[i]
print(w)
}
On 01-02-2012, at 16:55, Chris82 wrote:
Hi R users,
is there any possibilty that a while loop is working like that:
z - c(0,1,2,3,4,5,6,7,8,9)
r - 7
while(w == T) {
for ( i in 1:10 ){
w - r == z[i]
print(w)
}
}
The loop should
Thanks to both.
This is just a simple example. In real I have two vectors with different
lengths.
The code consists of two for loops for r and z. The main problem is the
computational time, so I try to stop the loop if w is TRUE for the first
time.
First I tried to use the command stop in
No. The while loop is only tested after the for loop has completed. Use debug
to understand this if it doesn't make sense to you.
---
Jeff NewmillerThe . . Go Live...
I guess you want something like
w=F
z - c(0,1,2,3,4,5,6,7,8,9)
r - 7
while(w == F) {
for ( i in 1:10 ){
w - r == z[i]
print(w)
}
}
But this loop will run forever. The condition for while is checked
when i jumps from 10 to 1. At that moment w
On 01-02-2012, at 17:32, Chris82 wrote:
Thanks to both.
This is just a simple example. In real I have two vectors with different
lengths.
The code consists of two for loops for r and z. The main problem is the
computational time, so I try to stop the loop if w is TRUE for the first
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