Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
Hi
wootten.adrie...@gmail.com napsal dne 12.08.2010 14:15:30:
Not quite what I was trying to say. The process generates a random
uniform
number between 0 and 1 and compares to a specific conditional
probability. It
is looking for this in particular:
random number Pr( rain(station=i,day=d)=1 | rain(station=i,day=d-1)=0
rain
(station=j,day=d)=0 rain(station=k,day=d)=0)
In this particular example, if the random number is less than the
probability
the value for station i and day d will be given as 1, otherwise it will
be zero.
There are 8 possible combinations. i is the station to be generated, j
and k
are the two stations most strongly correlated with station i. Stations
j and
k have already been generated in the example I gave previously. So I
want to
know given what is going on at stations j and k during day d and at
station i
for day d-1 if the value for station i day d will be 1 or 0.
But AFAIK that is what I said. Did you try anything from what I suggested?
You have 4 possible combinations from 2 stations (I named them col1 col2,
but you can name them differently - station j and k). So vector named cols
results in numbers 1:4 based on values col1 and col2 and you can do it
outside of loop.
If col1 and col2 are 0/1 vectors you can get it e.g. by
cols - (col1+(col2+1)*2)-1
and you get vector based on 0/1 value combination of your 2 vectors
(please try it:-)
You have 8 combinations of probabilities outcome (you call it specific
probability) and I presume it is a vector of 8 values (you still does not
provide small ***reproducible*** example)
So you can generate vector of random numbers with outside of loop and
compute combination of all possible outcomes with
ran-runif(20)
p-runif(8)
comparison - outer(ran,p, )
You will get comparison matrix e.g. for one month and you can just choose
appropriate row/column value in cycle based on cols value and previous day
value in station i.
Hope this provides some clarification.
A
If you do not provide some data with input and required specific outcome
you won't get specific answer. Instead of trying to explain it by
elaborated text use dput for exporting objects needed for computation and
if possible also the outcome.
Regards
Petr
On Thu, Aug 12, 2010 at 3:21 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
without toy example it is rather complicated to check your function. So
only few remarks:
Instead of generating 1 random number inside a loop generate whole
vector
of random numbers outside a loop and choose a number
Do not mix ifelse with if. ifelse is intended to work with whole vector.
Work with matrices instead of data frames whenever possible if speed is
an
issue.
If I understand correctly you want to put 1 or 0 into one column based
on:
previous value in the same column
comparison of some random number with predefined probabilities in vector
of 6 values
So here is vectorised version of your 4 ifs based on assumption
0 in col1 0 in col 2 = 5
0 in col1 1 in col 2 = 9
1 in col1 0 in col 2 = 6
1 in col1 1 in col 2 =10
col1-sample(1:2, 20, replace=T)
col2-sample(c(4,8), 20, replace=T)
col1+col2
[1] 5 6 9 6 6 5 9 10 9 9 6 9 10 6 10 9 10 9 5 5
cols-as.numeric(as.factor(col1+col2))
cols
[1] 1 2 3 2 2 1 3 4 3 3 2 3 4 2 4 3 4 3 1 1
And here is computed comparison of six values p (ortho obs used) with 20
generated random values
ran-runif(20)
p-runif(8)
comparison - outer(ran,p, )
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[2,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[3,] FALSE TRUE FALSE TRUE FALSE TRUE TRUE FALSE
[4,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[5,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[6,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[7,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[8,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[9,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[10,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[11,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[12,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[13,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[14,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[15,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[16,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[17,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[18,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[19,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[20,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
Now the only what you need to put in loop is to select appropriate
column
from matrix comparison based on value on cols vector and 0 or 1 in
previous row of station column.
Something like (untested)
gen.log-rep(NA, nrow(genmat)-1)
for (i in 2:nrow(genmat)) {
gen.log[i] - if( genmat[i-1, num] ==0)
Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
I did take your advice and change a few things in it to help it run. After reading through your earlier reply again I understood exactly what you were saying, so I did apply it in my function too. Thanks for all the advice! I appreciate it! Adrienne On Fri, Aug 13, 2010 at 3:29 AM, Petr PIKAL petr.pi...@precheza.cz wrote: Hi wootten.adrie...@gmail.com napsal dne 12.08.2010 14:15:30: Not quite what I was trying to say. The process generates a random uniform number between 0 and 1 and compares to a specific conditional probability. It is looking for this in particular: random number Pr( rain(station=i,day=d)=1 | rain(station=i,day=d-1)=0 rain (station=j,day=d)=0 rain(station=k,day=d)=0) In this particular example, if the random number is less than the probability the value for station i and day d will be given as 1, otherwise it will be zero. There are 8 possible combinations. i is the station to be generated, j and k are the two stations most strongly correlated with station i. Stations j and k have already been generated in the example I gave previously. So I want to know given what is going on at stations j and k during day d and at station i for day d-1 if the value for station i day d will be 1 or 0. But AFAIK that is what I said. Did you try anything from what I suggested? You have 4 possible combinations from 2 stations (I named them col1 col2, but you can name them differently - station j and k). So vector named cols results in numbers 1:4 based on values col1 and col2 and you can do it outside of loop. If col1 and col2 are 0/1 vectors you can get it e.g. by cols - (col1+(col2+1)*2)-1 and you get vector based on 0/1 value combination of your 2 vectors (please try it:-) You have 8 combinations of probabilities outcome (you call it specific probability) and I presume it is a vector of 8 values (you still does not provide small ***reproducible*** example) So you can generate vector of random numbers with outside of loop and compute combination of all possible outcomes with ran-runif(20) p-runif(8) comparison - outer(ran,p, ) You will get comparison matrix e.g. for one month and you can just choose appropriate row/column value in cycle based on cols value and previous day value in station i. Hope this provides some clarification. A If you do not provide some data with input and required specific outcome you won't get specific answer. Instead of trying to explain it by elaborated text use dput for exporting objects needed for computation and if possible also the outcome. Regards Petr On Thu, Aug 12, 2010 at 3:21 AM, Petr PIKAL petr.pi...@precheza.cz wrote: Hi without toy example it is rather complicated to check your function. So only few remarks: Instead of generating 1 random number inside a loop generate whole vector of random numbers outside a loop and choose a number Do not mix ifelse with if. ifelse is intended to work with whole vector. Work with matrices instead of data frames whenever possible if speed is an issue. If I understand correctly you want to put 1 or 0 into one column based on: previous value in the same column comparison of some random number with predefined probabilities in vector of 6 values So here is vectorised version of your 4 ifs based on assumption 0 in col1 0 in col 2 = 5 0 in col1 1 in col 2 = 9 1 in col1 0 in col 2 = 6 1 in col1 1 in col 2 =10 col1-sample(1:2, 20, replace=T) col2-sample(c(4,8), 20, replace=T) col1+col2 [1] 5 6 9 6 6 5 9 10 9 9 6 9 10 6 10 9 10 9 5 5 cols-as.numeric(as.factor(col1+col2)) cols [1] 1 2 3 2 2 1 3 4 3 3 2 3 4 2 4 3 4 3 1 1 And here is computed comparison of six values p (ortho obs used) with 20 generated random values ran-runif(20) p-runif(8) comparison - outer(ran,p, ) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [2,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE [3,] FALSE TRUE FALSE TRUE FALSE TRUE TRUE FALSE [4,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE [5,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE [6,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE [7,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE [8,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [9,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [10,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE [11,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE [12,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [13,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE [14,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [15,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE [16,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [17,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE [18,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE [19,] FALSE TRUE FALSE TRUE TRUE
Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
Not quite what I was trying to say. The process generates a random uniform
number between 0 and 1 and compares to a specific conditional probability.
It is looking for this in particular:
random number Pr( rain(station=i,day=d)=1 | rain(station=i,day=d-1)=0
rain(station=j,day=d)=0 rain(station=k,day=d)=0)
In this particular example, if the random number is less than the
probability the value for station i and day d will be given as 1, otherwise
it will be zero.
There are 8 possible combinations. i is the station to be generated, j and
k are the two stations most strongly correlated with station i. Stations j
and k have already been generated in the example I gave previously. So I
want to know given what is going on at stations j and k during day d and at
station i for day d-1 if the value for station i day d will be 1 or 0.
Hope this provides some clarification.
A
On Thu, Aug 12, 2010 at 3:21 AM, Petr PIKAL petr.pi...@precheza.cz wrote:
Hi
without toy example it is rather complicated to check your function. So
only few remarks:
Instead of generating 1 random number inside a loop generate whole vector
of random numbers outside a loop and choose a number
Do not mix ifelse with if. ifelse is intended to work with whole vector.
Work with matrices instead of data frames whenever possible if speed is an
issue.
If I understand correctly you want to put 1 or 0 into one column based on:
previous value in the same column
comparison of some random number with predefined probabilities in vector
of 6 values
So here is vectorised version of your 4 ifs based on assumption
0 in col1 0 in col 2 = 5
0 in col1 1 in col 2 = 9
1 in col1 0 in col 2 = 6
1 in col1 1 in col 2 =10
col1-sample(1:2, 20, replace=T)
col2-sample(c(4,8), 20, replace=T)
col1+col2
[1] 5 6 9 6 6 5 9 10 9 9 6 9 10 6 10 9 10 9 5 5
cols-as.numeric(as.factor(col1+col2))
cols
[1] 1 2 3 2 2 1 3 4 3 3 2 3 4 2 4 3 4 3 1 1
And here is computed comparison of six values p (ortho obs used) with 20
generated random values
ran-runif(20)
p-runif(8)
comparison - outer(ran,p, )
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[2,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[3,] FALSE TRUE FALSE TRUE FALSE TRUE TRUE FALSE
[4,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[5,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[6,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[7,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[8,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[9,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[10,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[11,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[12,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[13,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[14,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[15,] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
[16,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[17,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[18,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[19,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[20,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
Now the only what you need to put in loop is to select appropriate column
from matrix comparison based on value on cols vector and 0 or 1 in
previous row of station column.
Something like (untested)
gen.log-rep(NA, nrow(genmat)-1)
for (i in 2:nrow(genmat)) {
gen.log[i] - if( genmat[i-1, num] ==0) comparison[i, cols[i]] else
comparison[i,cols[i+5]]
}
genmat[2:nrow(genmat), num] - gen.log*1
Regards
Petr
r-help-boun...@r-project.org napsal dne 11.08.2010 18:35:37:
Hello Everyone!
Here's what I'm trying to do. I'm working on generating occurrences of
precipitation based upon precipitation occurrence for a station during
the
previous day and two stations that have already been generated by joint
probablities and 1st order Markov chains or by the same generation
process.
This has to be done for each remaining stations for each month.
genmat # 7 stations in this example, line_before is the climatology of
the
last day of the previous month. Stations 4 and 6 have been generated
already
in this example
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
line_before1110111
NA NA NA1 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0
Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
Thanks everyone for your help and advice. For the R-help archives, here is
what I ended up doing.
First creating a separate function to handle one day at a time -
byrow.gen2 - function(genmat,rownum,use1,use2,num,ortho_obs_used){
prev = rownum-1
ran = runif(length(rownum),0,1)
if(genmat[rownum,use1]==0 genmat[rownum,use2]==0 genmat[prev,num]==0) {
if(ranortho_obs_used$Pr[1]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==0 genmat[rownum,use2]==0 genmat[prev,num]==1) {
if(ranortho_obs_used$Pr[4]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==0 genmat[rownum,use2]==1 genmat[prev,num]==0) {
if(ranortho_obs_used$Pr[2]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==0 genmat[rownum,use2]==1 genmat[prev,num]==1) {
if(ranortho_obs_used$Pr[5]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==1 genmat[rownum,use2]==0 genmat[prev,num]==0) {
if(ranortho_obs_used$Pr[3]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==1 genmat[rownum,use2]==0 genmat[prev,num]==1) {
if(ranortho_obs_used$Pr[7]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==1 genmat[rownum,use2]==1 genmat[prev,num]==0) {
if(ranortho_obs_used$Pr[6]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
if(genmat[rownum,use1]==1 genmat[rownum,use2]==1 genmat[prev,num]==1) {
if(ranortho_obs_used$Pr[8]){ genmat[rownum,num] = 1 }else{
genmat[rownum,num] = 0}
}
genmat
}
Then applying the foreach package in the original function
event.gen3 = function(genmat,use1,use2,num,ortho_obs_used){
rownum = 2:nrow(genmat)
test = foreach(r=iter(rownum,by='row')) %dopar% { genmat =
byrow.gen2(genmat,r,use1,use2,num,ortho_obs_used) }
rm(test)
genmat
}
The final results were exactly as I needed them to be in my initial post,
but the processing time dropped from 2 seconds per station to 0.05 seconds
per station.
Thanks to everyone for giving me the advice and the idea to try this!
Adrienne
On Thu, Aug 12, 2010 at 8:15 AM, Adrienne Wootten amwoo...@ncsu.edu wrote:
Not quite what I was trying to say. The process generates a random uniform
number between 0 and 1 and compares to a specific conditional probability.
It is looking for this in particular:
random number Pr( rain(station=i,day=d)=1 | rain(station=i,day=d-1)=0
rain(station=j,day=d)=0 rain(station=k,day=d)=0)
In this particular example, if the random number is less than the
probability the value for station i and day d will be given as 1, otherwise
it will be zero.
There are 8 possible combinations. i is the station to be generated, j and
k are the two stations most strongly correlated with station i. Stations j
and k have already been generated in the example I gave previously. So I
want to know given what is going on at stations j and k during day d and at
station i for day d-1 if the value for station i day d will be 1 or 0.
Hope this provides some clarification.
A
On Thu, Aug 12, 2010 at 3:21 AM, Petr PIKAL petr.pi...@precheza.czwrote:
Hi
without toy example it is rather complicated to check your function. So
only few remarks:
Instead of generating 1 random number inside a loop generate whole vector
of random numbers outside a loop and choose a number
Do not mix ifelse with if. ifelse is intended to work with whole vector.
Work with matrices instead of data frames whenever possible if speed is an
issue.
If I understand correctly you want to put 1 or 0 into one column based on:
previous value in the same column
comparison of some random number with predefined probabilities in vector
of 6 values
So here is vectorised version of your 4 ifs based on assumption
0 in col1 0 in col 2 = 5
0 in col1 1 in col 2 = 9
1 in col1 0 in col 2 = 6
1 in col1 1 in col 2 =10
col1-sample(1:2, 20, replace=T)
col2-sample(c(4,8), 20, replace=T)
col1+col2
[1] 5 6 9 6 6 5 9 10 9 9 6 9 10 6 10 9 10 9 5 5
cols-as.numeric(as.factor(col1+col2))
cols
[1] 1 2 3 2 2 1 3 4 3 3 2 3 4 2 4 3 4 3 1 1
And here is computed comparison of six values p (ortho obs used) with 20
generated random values
ran-runif(20)
p-runif(8)
comparison - outer(ran,p, )
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[2,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[3,] FALSE TRUE FALSE TRUE FALSE TRUE TRUE FALSE
[4,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[5,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[6,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[7,] FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE
[8,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[9,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[10,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[11,] FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
[12,] FALSE TRUE FALSE TRUE TRUE TRUE TRUE TRUE
[R] Running something without a loop when the result from the previous iteration is require for the current iteration
Hello Everyone!
Here's what I'm trying to do. I'm working on generating occurrences of
precipitation based upon precipitation occurrence for a station during the
previous day and two stations that have already been generated by joint
probablities and 1st order Markov chains or by the same generation process.
This has to be done for each remaining stations for each month.
genmat # 7 stations in this example, line_before is the climatology of the
last day of the previous month. Stations 4 and 6 have been generated already
in this example
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
line_before1110111
NA NA NA1 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
num # station to generate
[1] 2
use1 # 1st station to use in generation
[1] 6
use2 # 2nd station to use in generation
[1] 4
genmat = event.gen2(genmat,use1,use2,num,ortho_obs_used) # Generation
function shown below
genmat # genmat - after it has gone through station 2
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
line_before1110111
NA0 NA1 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA1 NA0 NA
NA1 NA1 NA1 NA
NA1 NA1 NA1 NA
NA1 NA0 NA0 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA0 NA1 NA1 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
NA1 NA0 NA1 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA1 NA1 NA1 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
Where event.gen2 is this function:
event.gen2 = function(genmat,use1,use2,num,ortho_obs_used){
for(r in 2:nrow(genmat)){
ran = runif(1,0,1)
if(genmat[r,use1]==0 genmat[r,use2]==0){
genmat[r,num]-ifelse(genmat[r-1,num]==0,ifelse(ranortho_obs_used$Pr[1],1,0),ifelse(ranortho_obs_used$Pr[4],1,0))
}
if(genmat[r,use1]==0 genmat[r,use2]==1){
genmat[r,num]-ifelse(genmat[r-1,num]==0,ifelse(ranortho_obs_used$Pr[2],1,0),ifelse(ranortho_obs_used$Pr[5],1,0))
}
if(genmat[r,use1]==1 genmat[r,use2]==0){
genmat[r,num]-ifelse(genmat[r-1,num]==0,ifelse(ranortho_obs_used$Pr[3],1,0),ifelse(ranortho_obs_used$Pr[7],1,0))
}
if(genmat[r,use1]==1 genmat[r,use2]==1){
genmat[r,num]-ifelse(genmat[r-1,num]==0,ifelse(ranortho_obs_used$Pr[6],1,0),ifelse(ranortho_obs_used$Pr[8],1,0))
}
gc()
}
genmat
}
ortho_obs_used is a data frame that contains the probablity of precipitation
occurring on a given day for a specific set of condtions.
For instance ortho_obs_used$Pr[1] is the probablity of rain at station s for
day d, given that there was no rain at station s for day d-1
Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
Hi Adrienne,
I guess apply should be better than for loop.
Code like this:
event.gen2 = function(genmat,use1,use2,num,ortho_obs_used){
onerow.gen - function(one.row, use1){
one.row[num] - ifelse(...}
genmat[,num] - NA ##Add one row with NA values
apply(genmat[-1,],1,onerow.gen,use1=...)}
And R looks TRUE as 1 and FALSE as 0, we may take advantage of it.
Regards,
Wu
-
A R learner.
--
View this message in context:
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Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
What is wrong with using a loop?
It used to be that loops were much slower than some of the alternatives, but
now days a well crafted loop runs almost as fast (sometime faster) than the
apply functions. So if the loop is working for you, use it and don't worry
about it (though there may be ways to speed up the loop).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Adrienne Wootten
Sent: Wednesday, August 11, 2010 10:36 AM
To: r-help@r-project.org
Subject: [R] Running something without a loop when the result from the
previous iteration is require for the current iteration
Hello Everyone!
Here's what I'm trying to do. I'm working on generating occurrences of
precipitation based upon precipitation occurrence for a station during
the
previous day and two stations that have already been generated by joint
probablities and 1st order Markov chains or by the same generation
process.
This has to be done for each remaining stations for each month.
genmat # 7 stations in this example, line_before is the climatology
of the
last day of the previous month. Stations 4 and 6 have been generated
already
in this example
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
line_before1110111
NA NA NA1 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA1 NA
NA NA NA0 NA0 NA
NA NA NA0 NA0 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA1 NA1 NA
NA NA NA0 NA0 NA
num # station to generate
[1] 2
use1 # 1st station to use in generation
[1] 6
use2 # 2nd station to use in generation
[1] 4
genmat = event.gen2(genmat,use1,use2,num,ortho_obs_used) # Generation
function shown below
genmat # genmat - after it has gone through station 2
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
line_before1110111
NA0 NA1 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA1 NA0 NA
NA1 NA1 NA1 NA
NA1 NA1 NA1 NA
NA1 NA0 NA0 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA0 NA1 NA1 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
NA1 NA0 NA1 NA
NA0 NA0 NA0 NA
NA0 NA0 NA0 NA
NA1 NA1 NA1 NA
NA1 NA1 NA1 NA
NA1 NA1 NA1 NA
NA0 NA0 NA0 NA
Where event.gen2 is this function:
event.gen2 = function(genmat,use1,use2,num,ortho_obs_used){
for(r in 2:nrow(genmat)){
ran = runif(1,0,1)
if(genmat[r,use1]==0 genmat[r,use2]==0){
Re: [R] Running something without a loop when the result from the previous iteration is require for the current iteration
If it were just one loop by itself and I was doing the calculation for just one month for all of my 317 stations, I would agree with. However, this function itself is inside another loop which goes through each month and year that I need the calculation for each of the stations. If you have any suggestions for how I could speed up the loop that is welcome, but I would like to try to remove it given that it is nested inside another loop. A On Wed, Aug 11, 2010 at 2:59 PM, Greg Snow greg.s...@imail.org wrote: What is wrong with using a loop? It used to be that loops were much slower than some of the alternatives, but now days a well crafted loop runs almost as fast (sometime faster) than the apply functions. So if the loop is working for you, use it and don't worry about it (though there may be ways to speed up the loop). -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Adrienne Wootten Sent: Wednesday, August 11, 2010 10:36 AM To: r-help@r-project.org Subject: [R] Running something without a loop when the result from the previous iteration is require for the current iteration Hello Everyone! Here's what I'm trying to do. I'm working on generating occurrences of precipitation based upon precipitation occurrence for a station during the previous day and two stations that have already been generated by joint probablities and 1st order Markov chains or by the same generation process. This has to be done for each remaining stations for each month. genmat # 7 stations in this example, line_before is the climatology of the last day of the previous month. Stations 4 and 6 have been generated already in this example [,1] [,2] [,3] [,4] [,5] [,6] [,7] line_before1110111 NA NA NA1 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA1 NA0 NA NA NA NA1 NA1 NA NA NA NA1 NA1 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA1 NA1 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA1 NA1 NA NA NA NA0 NA0 NA NA NA NA1 NA1 NA NA NA NA1 NA1 NA NA NA NA1 NA1 NA NA NA NA0 NA0 NA NA NA NA0 NA1 NA NA NA NA0 NA0 NA NA NA NA0 NA0 NA NA NA NA1 NA1 NA NA NA NA1 NA1 NA NA NA NA1 NA1 NA NA NA NA0 NA0 NA num # station to generate [1] 2 use1 # 1st station to use in generation [1] 6 use2 # 2nd station to use in generation [1] 4 genmat = event.gen2(genmat,use1,use2,num,ortho_obs_used) # Generation function shown below genmat # genmat - after it has gone through station 2 [,1] [,2] [,3] [,4] [,5] [,6] [,7] line_before1110111 NA0 NA1 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA1 NA0 NA NA1 NA1 NA1 NA NA1 NA1 NA1 NA NA1 NA0 NA0 NA NA0 NA0 NA0 NA NA1 NA1 NA1 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA0 NA0 NA0 NA NA1 NA1 NA1 NA NA0 NA0 NA0 NA NA1 NA1 NA1 NA NA0 NA1 NA1 NA NA1 NA
