Re: [R] _: how to replace values in x by means in subgroups created in ...(not loops)
Re:_: how to replace values in x by means in subgroups created in ...(not loops)
Thanks, below some code and reply:
#_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_#
#---
# my slow function with loops:
# replace_x_by_locallyMean_x_4_0s_in_y.f(x,y)
#---
#---
# replace_x_by_locallyMean_x_4_0s_in_y.f
# Arguments: function takes vectors x, y
#these vector have the same length: length(x)==length(y)
# function replaces in x
# values that have in indexes in y
# at least one zero in the same position in y
# by locally compted mean of values in x
#from all local x that also have continous 000 in their positions
close to this x
# in y vector
#x # 0 1 2 3 4 5 6 7 8 9
#y # 0 0 0 1 0 0 0 1 1 1
# Value expected:
#x_m # 1.5 1.5 1.5 3.0 5.0 5.0 5.0 7.0 8.0 9.0
#---
# author: me
# replace_x_by_locallyMean_x_4_0s_in_y.f(x,y)
#---
#initialisation
Dff-NULL
index_Dff.start_0-NULL
index_Dff.stop_0-NULL
x-NULL
y-NULL
replace_x_by_locallyMean_x_4_0s_in_y.f-NULL
replace_x_by_locallyMean_x_4_0s_in_y.f-function(x=Df$x,
y=Df$y){
Dff-as.data.frame(cbind(as.vector(x),as.vector(y)))
colnames(Dff)-c('x','y')
index_Dff.start_0-which( ( (Dff$y==0)
rev(rev(c(0,Dff$y))[-1]) !=0) )
if (Dff$y[1]==0) {
index_Dff.start_0=c(1,index_Dff.start_0)
}
index_Dff.stop_0-which( ( (Dff$y==0)
c(Dff$y[-1],rev(Dff$y)[1]) !=0) )
if (rev(Dff$y)[1]==0) {
index_Dff.stop_0=c(index_Dff.stop_0,length(x))
}
Dff$x_m-Dff$x
for (i in ( 1:length(index_Dff.start_0) ) ) {
Dff$x_m[index_Dff.start_0[i]:index_Dff.stop_0[i]]-mean(
Dff$x[index_Dff.start_0[i]:index_Dff.stop_0[i]],na.rm=T)
}
for (i in ( 1:length(index_Dff.start_0) ) ) {
Dff$x_m[index_Dff.start_0[i]:index_Dff.stop_0[i]]-mean(
Dff$x[index_Dff.start_0[i]:index_Dff.stop_0[i]],na.rm=T)
}
Dff$x_m
}
#replace_x_by_locallyMean_x_4_0s_in_y.f()
#---
# replace_x_by_locallyMean_x_4_0s_in_y.f__Thierry
# Arguments: function takes vectors x, y
#these vector have the same length: length(x)==length(y)
# function replaces in x
# values that have in indexes in y
# at least one zero in the same position in y
# by locally compted mean of values in x
#from all local x that also have continous 000 in their positions
close to this x
# in y vector
#x # 0 1 2 3 4 5 6 7 8 9
#y # 0 0 0 1 0 0 0 1 1 1
# Value expected:
#x_m # 1.5 1.5 1.5 3.0 5.0 5.0 5.0 7.0 8.0 9.0
#---
# author: Thierry
# replace_x_by_Avgx_for000iny_Thierry.f(x,y)
#---
#initialisation
dataset - NULL
tmp - NULL
result - NULL
replace_x_by_locallyMean_x_4_0s_in_y.f__Thierry-NULL
library(plyr)
replace_x_by_locallyMean_x_4_0s_in_y.f__Thierry-function(x=x,
y=y){
dataset - data.frame(x = x, y = y)
dataset$Group - cumsum(c(0, diff(!is.na(dataset$y) dataset$y == 0)) == 1)
#library(plyr)
tmp - ddply(subset(dataset, y == 0), .(Group), function(z){c(Mean =
mean(z$x, na.rm = TRUE))})
result - merge(dataset, tmp)
result$Mean[is.na(result$y) | result$y != 0] - result$x[is.na(result$y) |
result$y != 0]
#result
result$Mean
} #replace_x_by_locallyMean_x_4_0s_in_y.f__Thierry(x,y)
# Let's test it for longer vector
x-1:(50*1000)
y-sample(c(0,1), length(x), replace = TRUE)
# for Thierry we must start by 1 if it start by 0,
y-c(0,1,sample(c(0,1), length(x)-2, replace = TRUE) )
paste(length(x),length(y))
# let's comapre times:
# slow solution with loops, my function x_by_locallyMean_x_4_0s_in_y__loop()
# system.time(
# x_by_locallyMean_x_4_0s_in_y__loop-
#replace_x_by_locallyMean_x_4_0s_in_y.f(x,y)
# )
# _for_ x-1:(50*1000) #
# user system elapsed
# 760.340.86 769.16
#---
# time of Thierry's function :
system.time(
x_by_locallyMean_x_4_0s_in_y__Thierry-
replace_x_by_locallyMean_x_4_0s_in_y.f__Thierry(x,y)
)
#Df.x_m_Thierry (10x faster) then my loop
# _for_ x-1:50
# user system elapsed
# 11.610.00 11.68
# good news:
# 769.16 sec / 11.68 sec = Thierry's solution is 65 (x) times faster than
loops,
# great!
length(x_by_locallyMean_x_4_0s_in_y__Thierry)
cbind(x,y,x_by_locallyMean_x_4_0s_in_y__Thierry) [1:15,]
paste(' good, faster than my loops ')
#1. Thierry, your function is fast, ()
[R] how to replace values in x by means in subgroups created in ... (not loops)
# Dear all, # how to replace values in x by means in subgroups created in ... # replace only these values where y=0 in continous sequence # replace by mean calculated locally for each subgroup created by # continous sequence of 0,0,0 in parallel y vector, i.e. # where there is continous sequence of 0 in data frame vector y #but we do not replace values in x[i], if y[i]!=0 # we do not want use loops we do not use apply (not very fast) x -c(0 ,1,2,3,4,5,6,7,8,NA,NA,1 ,1 ,NA,2 ,2) y -c(0 ,0,0,1,0,0,0,1,1,1 ,NA,0 ,0 ,0 ,0 ,1) Must_be-c(1 ,1,1,3,5,5,5,7,8,NA,NA,1.5,1 ,NA,1.5,2) (df-as.data.frame(cbind(x,y)) ) # I have traied many bad colusions based on cumsum, pmin, pmax, ... (mean_dfx_if_yIs0-y*cumsum(x*y)/(cumsum(y)*y) ) # how to do this? # thans for any advice # E [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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.
Re: [R] how to replace values in x by means in subgroups created in ... (not loops)
Something like this?
dataset - data.frame(x = x, y = y)
dataset$Group - cumsum(c(0, diff(!is.na(dataset$y) dataset$y == 0)) == 1)
library(plyr)
tmp - ddply(subset(dataset, y == 0), .(Group), function(z){c(Mean = mean(z$x,
na.rm = TRUE))})
result - merge(dataset, tmp)
result$Mean[is.na(result$y) | result$y != 0] - result$x[is.na(result$y) |
result$y != 0]
Best regards,
Thierry
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Eugeniusz Kaluza
Verzonden: woensdag 27 juli 2011 11:21
Aan: r-help@r-project.org
Onderwerp: [R] how to replace values in x by means in subgroups created in ...
(not loops)
# Dear all,
# how to replace values in x by means in subgroups created in ...
# replace only these values where y=0 in continous sequence # replace by mean
calculated locally for each subgroup created by # continous sequence of 0,0,0
in
parallel y vector, i.e.
# where there is continous sequence of 0 in data frame vector y
#but we do not replace values in x[i], if y[i]!=0
# we do not want use loops we do not use apply (not very fast)
x -c(0 ,1,2,3,4,5,6,7,8,NA,NA,1 ,1 ,NA,2 ,2)
y -c(0 ,0,0,1,0,0,0,1,1,1 ,NA,0 ,0 ,0 ,0 ,1)
Must_be-c(1 ,1,1,3,5,5,5,7,8,NA,NA,1.5,1 ,NA,1.5,2)
(df-as.data.frame(cbind(x,y)) )
# I have traied many bad colusions based on cumsum, pmin, pmax, ...
(mean_dfx_if_yIs0-y*cumsum(x*y)/(cumsum(y)*y) )
# how to do this?
# thans for any advice
# E
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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@r-project.org mailing list
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.
