[R] filter and merge 2 big Df's by=t_chr in sec.
Please help to solve that problem: filter and merge 2 big Df's by=t_chr in sec. require(zoo) t1=2012-03-25 17:00:00.0 t2=2012-03-25 17:00:05.0 t1_POSIXlt-strptime(t1, %Y-%m-%d %H:%M:%OS)# t2_POSIXlt-strptime(t2, %Y-%m-%d %H:%M:%OS)# # ts_t1t2_POSIXlt-seq(t1_POSIXlt,t2_POSIXlt,length.out =-1+2*length(seq(t1_POSIXlt,t2_POSIXlt, sec) ) ) ts_t1t2_POSIXlt-seq(t1_POSIXlt,t2_POSIXlt,length.out =length(seq(t1_POSIXlt,t2_POSIXlt, sec) ) ) t_ts_t1t2_chr-as.character(format(ts_t1t2_POSIXlt,%Y-%m-%d %H:%M:%OS)) tail(t_ts_t1t2_chr) t_chr-rep(t_ts_t1t2_chr,c((1:(-1+length(t_ts_t1t2_chr))),1)) str(t_chr) # system.time() Df.a-as.data.frame(cbind(t_chr,a=c(1:length(t_chr)),b=2*c(1:length(t_chr)),c=0.1*c(length(t_chr):1))); Df.a # #We wants to obtain for each single second (time as a key) a) mean filtered t_chr a b c 2012-03-25 17:00:00 1 2 1.6 2012-03-25 17:00:01 2.5 5 1.45 2012-03-25 17:00:02 5 101.2 2012-03-25 17:00:03 8.5 170.85 2012-03-25 17:00:04 13 26 0 0.4 2012-03-25 17:00:05 16 320.1 Df.a.filtered.mean=??? #-- #We wants to obtain for each single second (time as a key) b) max filtered #key = time, each second: 1 max from these same seconds t_chr a b c 1 2012-03-25 17:00:00 1 2 1.6 3 2012-03-25 17:00:01 3 6 1.5 6 2012-03-25 17:00:02 6 12 1.3 10 2012-03-25 17:00:03 10 20 1.0 10 2012-03-25 17:00:04 15 30 1.0 16 2012-03-25 17:00:05 16 32 0.1 Df.a.filtered.max =??? #-- the problem is, taht we have huge matrix. The solution should be fast. next we want to merge it with another Data frame #-- t1=2012-03-25 17:00:00.0 t2=2012-03-25 17:00:05.0 t1_POSIXlt-strptime(t1, %Y-%m-%d %H:%M:%OS)# t2_POSIXlt-strptime(t2, %Y-%m-%d %H:%M:%OS)# ts_t1t2_POSIXlt-seq(t1_POSIXlt,t2_POSIXlt,length.out =-1+2*length(seq(t1_POSIXlt,t2_POSIXlt, sec) ) ) # ts_t1t2_POSIXlt-seq(t1_POSIXlt,t2_POSIXlt,length.out =length(seq(t1_POSIXlt,t2_POSIXlt, sec) ) ) t_chr2-t_ts_t1t2_chr-as.character(format(ts_t1t2_POSIXlt,%Y-%m-%d %H:%M:%OS)) tail(t_ts_t1t2_chr) Df.b-as.data.frame(cbind(t_chr2,dfb.b1=c(1:length(t_chr2)),dfb.b=2*c(1:length(t_chr2)),dfb.c=0.1*c(length(t_chr2):1))); Df.b #-- merge(Df.a,Df.b, by.x=t_chr,by.y=t_chr2,all.x=TRUE,all.y=TRUE) And if in Df.a there is no all seconds measurements? And if in Df.a. there are many NA? How we can handle it in R? Thanks KaluzaEA. [[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)
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.
[R] How to filter XY pairs of inacurate gps position along track, taking into account the time index to not mix track from different days in one average track
Dear R users,
subject: How to filter XY pairs of inacurate gps position along track, taking
into account the time index to not mix track from different days in one average
track
or subject as: How to filter XY pairs of inacurate gps position: (latitude,
longtitude)
to get estimated correct position before calculating track
i.e.
How to estimate (time series? )?
How to filter XY pairs of inacurate gps position along track:
Lon-c(18.13015167,18.13022667,18.13028500,18.13035500,18.13041333,18.13047000,18.13054667,18.13060333,18.13067500,18.13073167,18.13079000,18.13085167,18.13091000,18.13096667,18.13103000,18.13109500,18.13114500,18.13120167,18.13124667,18.13130500,18.13136167,18.13140667,18.13146333,18.13150833,18.13156500,18.13161000,18.13167167,18.13171667,18.13177000,18.13181333,18.13187167,18.13191667,18.13197333,18.13203167,18.13208833,18.13214667,18.13220333,18.13225333,18.13231000,18.13236833,18.13243000,18.13249167,18.13254167,18.13259833,18.13265667,18.13270500,18.13276333,18.13282000,18.13287000,18.13292667,18.13298500,18.13304667,18.13310500,18.13315333,18.13321500,18.13327833,18.13332667)
Lat-c(48.95961500,48.95961667,48.95961500,48.95961000,48.95960667,48.95959833,48.95960167,48.95959833,48.95959333,48.95959167,48.95959000,48.95959333,48.95959167,48.95958833,48.95959333,48.95958167,48.95958833,48.95958500,48.95958500,48.95958333,48.95958000,48.95957500,48.95957167,48.95957167,48.95956500,48.95955833,48.95956833,48.95956833,48.95955833,48.95955333,48.95955167,48.95955000,48.95954333,48.95954000,48.95953833,48.95953500,48.9595,48.95954000,48.95953667,48.95954000,48.95954500,48.95954833,48.95955500,48.95955833,48.95955667,48.95956167,48.95956000,48.95955833,48.95957000,48.95956667,48.95956000,48.95956333,48.95956167,48.95956833,48.95957167,48.95958167,48.95958833)
t_index-1:(length(Lat))
Raw-as.data.frame(cbind(t_index,Lon,Lat))
colnames(Raw)-c('t_index','Lon','Lat')
plot(Raw$Lon,Raw$Lat,type=l,main = lowess(cars))
# data should be or filtared or ...?
#expected answer:
# ver 1
--
#Raw$Lon_v2- f_x (x=Raw$Lon, y=Raw$Lat)
#Raw$Lat_v2 - f_y (x=Raw$Lon, y=Raw$Lat)
#lines(Raw$Lon_v2,Raw$Lat_v2,col=2)
# ver 2
--
op - par(mfrow = c(2,1), mgp = c(2,.8,0), mar = .1+c(3,3,3,1))
plot(Raw$Lon,Raw$Lat,type=l,main = lowess())
f_k2=0.3
Raw$Lon_v2-(lowess(x=Raw$Lon, y=Raw$Lat,f=f_k2))$x
Raw$Lat_v2-(lowess(x=Raw$Lon, y=Raw$Lat,f=f_k2))$y
points(Raw$Lon_v2,Raw$Lat_v2,col=2,pch='+')
lines(Raw$Lon_v2,Raw$Lat_v2,col=2)
f_k3=0.6
Raw$Lon_v2-(lowess(x=Raw$Lon, y=Raw$Lat,f=f_k3))$x
Raw$Lat_v2-(lowess(x=Raw$Lon, y=Raw$Lat,f=f_k3))$y
points(Raw$Lon_v2,Raw$Lat_v2,col=3,pch='x')
lines(Raw$Lon_v2,Raw$Lat_v2,col=3,lwd=2)
legend(Raw$Lon_v2[1],Raw$Lat_v2[1], c(paste(f = ,
c(round(f_k2,2),round(f_k3,2, lty = 1, col = 2:3)
par(op)
How to approximate these positions along some possible track (line, curve)
for all localizations along the track, that can change directions.
The above methods are not good, because, if there are many track in vincinity
of the first track, the above method
will average the points between track, that are close each another.
And we need to aproximate only along one time track,
i.e. the car (Lat Lon position) in one moment must be laid along one route
(lane) (today's track)
and not along the route that it will cover 2 days ago, only because that is
another day, the another index of point.
Yours sincerely,
Eugen
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] create group markers in original data frame ie.countinued... ? to calculate sth for groups defined betweenpoints in one variable (string), /separating/ spliting variable into groups by i.e. be
Dear useRs, and expeRts,
tahanks
I have found idea how to add to oryginal data a column with markers to know
with all data in wchich period in c2 they are, suimply in the code I could add:
stacked_idx-stack(idx)
merge(stacked_idx,C.df,by.x=c('values'),by.y=c('c0'), all=T)
thanks for suggestions,
Kaluza
-Wiadomość oryginalna-
Od: r-help-boun...@r-project.org w imieniu Eugeniusz Kaluza
Wysłano: Pt 2010-06-25 14:48
Do: c
Temat: Re: [R] create group markers in original data frame ie.countinued... ?
to calculate sth for groups defined betweenpoints in one variable
(string),/separating/ spliting variable into groups by i.e. between A-B,B-C,
C-D, from: A, NA, NA, B, NA, NA, C, NA, NA, NA, D
Dear useRs,
at the beginning,
Joris Meys, thank you for explaining how to obtain calculation result possible
for groups between string marks in one variable in data frame, like in this
example below (between START and STOP), wchich I would like to complete at the
end by asking about... how is possible to mark each observations presented in
oryginal data set
# so firstly, below
# START...working example of solution proposed by: Joris Meys
[jorism...@gmail.com]
# Same trick :
c0-rbind( 1, 2 , 3, 4, 5, 6, 7, 8, 9,10,11,
12,13,14,15,16,17 )
c0
c1-rbind(10, 20 ,30,40, 50,10,60,20,30,40,50, 30,10,
0,NA,20,10.3444)
c1
c2-rbind(NA,A,NA,NA,B,NA,NA,NA,NA,NA,NA,C,NA,NA,NA,NA,D)
c2
pos - which(!is.na(C.df$c2))
idx - sapply(2:length(pos),function(i) pos[i-1]:(pos[i]-1))
names(idx) - sapply(2:length(pos),
function(i) paste(C.df$c2[pos[i-1]],-,C.df$c2[pos[i]]))
out - lapply(idx,function(i) summary(C.df[i,1:2]))
out
#STOP ... below from: Sent: Thu 2010-06-24 18:02: Joris Meys
[jorism...@gmail.com]
#Thank you, it is done and works very well
# - - - - - - - -- - - - - - -- - -
# Now, I try to finish my question to add gruping sybol to the whole set,
making
# each observation marked by the name of the interval in which that observation
is placed.
# to tell the observator, that this observation is between ...A and B, to
enable sorting, to eneable simple acess using match
in_sub_starting_from-rbind(NA,A,A,A,B,B,B,B,B,B,B,C,C,C,C,C,C)
in_sub_finished_by
-rbind(NA,B,B,B,C,C,C,C,C,C,C,D,D,D,D,D,D)
in_sub_limited_by-rbind(NA,A-B,A-B,A-B,B-C,B-C,B-C,B-C,B-C,B-C,B-C,C-D,C-D,C-D,C-D,C-D,C-D)
C.df-data.frame(c0,c1,c2,in_sub_starting_from,in_sub_finished_by,in_sub_limited_by)
C.df
#
# Therefore my one more question:
How is possible to create these vectors automaticly, having C.df$c2 (and of
course having also: C.df$c0,C.df$c1), :
C.df$in_sub_starting_from
C.df$in_sub_finished_by
C.df$in_sub_limited_by
#to tell the observator, that this observation is between ...A and B, to enable
sorting, to eneable simple acess using match
#for example, to make possible this access to data:
#to to take the 7'th observation from any row of data frame,
C.df$c0[7]
C.df$c1[c0==7]
#and could
#find in this same row in_sub_starting_from that observation is preceded by
...
C.df$in_sub_starting_from[c0==7]
#find in this same row in_sub_finished_by that observation is before ...
C.df$in_sub_finished_by[c0==7]
#find in this same row in_sub_finished_by that this observation is between ...
C.df$in_sub_limited_by[c0==7]
#
?
#Thanks for advices, and maybe and this answer,
#looking impatiently for time with possible access to internet...
#
Sincerely,
Kaluza
and the beginnig of this story;
-Original Message-
From: Eugeniusz Kaluza
Sent: Thu 2010-06-24 17:12
To: r-help@r-project.org
Subject: PD: [R] ?to calculate sth for groups defined between points in one
variable (string), / value separating/ spliting variable into groups by i.e.
between start, NA, NA, stop1, start2, NA, stop2
Dear useRs,
Thanks for advice from Joris Meys,
Now will try to think how to make it working for less specyfic case,
to make the problem more general.
Then the result should be displayed for every group between non empty string in
c2
i.e. not only result for:
#mean:
c1 c3c4 c5
20 Start1 Stop1 Start1-Stop1
25.48585 Start2 Stop2 Start2-Stop2
but also for every one group created by space between two closest strings in
c2, that contains only seriess of Na, NA, NA, separated from time to time by
one string i.e.:
#mean:
c1 c3c4 c5
20 Start1 Stop1 Start1-Stop1
.. Stop1 Start2 Stop1-Start2
25.48585 Start2 Stop2 Start2-Stop2
i.e.
to rewrite this maybe for another simpler version of command
but also for every one group created by space between two closest strings in
c2, that contains only seriess of Na, NA, NA, separated from time to time by
one
[R] PD: ?to calculate sth for groups defined between points in one variable (string), / value separating/ spliting variable into groups by i.e. between start, NA, NA, stop1, start2, NA, stop2
Dear useRs, Thanks for advice from Joris Meys, Now will try to think how to make it working for less specyfic case, to make the problem more general. Then the result should be displayed for every group between non empty string in c2 i.e. not only result for: #mean: c1 c3c4 c5 20 Start1 Stop1 Start1-Stop1 25.48585 Start2 Stop2 Start2-Stop2 but also for every one group created by space between two closest strings in c2, that contains only seriess of Na, NA, NA, separated from time to time by one string i.e.: #mean: c1 c3c4 c5 20 Start1 Stop1 Start1-Stop1 .. Stop1 Start2 Stop1-Start2 25.48585 Start2 Stop2 Start2-Stop2 i.e. to rewrite this maybe for another simpler version of command but also for every one group created by space between two closest strings in c2, that contains only seriess of Na, NA, NA, separated from time to time by one string A, NA, NA, NA, NA, B, NA, NA, NA, C, NA,NA,NA,NA,D, NA,NA i.e.: #mean: c1 c3c4 c5 20 A B A-B .. B C B-C 25.48585 C D C-D ... Looking for more general method (function), grouping between these letters in c2, I will now try to study solution proposed by Joris Meys Thanks for immediate aswer Kaluza -Wiadomo¶æ oryginalna- Od: Joris Meys [mailto:jorism...@gmail.com] Wys³ano: Cz 2010-06-24 15:14 Do: Eugeniusz Ka³u¿a DW: r-help@r-project.org Temat: Re: [R] ?to calculate sth for groups defined between points in one variable (string), / value separating/ spliting variable into groups by i.e. between start, NA, NA, stop1, start2, NA, stop2 On Thu, Jun 24, 2010 at 1:18 PM, Eugeniusz Kaluza eugeniusz.kal...@polsl.pl wrote: Dear useRs, Thanks for any advices # I do not know where are the examples how to mark groups # based on signal occurence in the additional variable: cf. variable c2, # How to calculate different calculations for groups defined by (split by occurence of c2 characteristic data) #First example of simple data #mexample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 c0-rbind( 1, 2 , 3, 4, 5, 6, 7, 8, 9,10,11, 12,13,14,15,16,17 ) c0 c1-rbind(10, 20 ,30,40, 50,10,60,20,30,40,50, 30,10, 0,NA,20,10.3444) c1 c2-rbind(NA,Start1,NA,NA,Stop1,NA,NA,NA,NA,NA,NA,Start2,NA,NA,NA,NA,Stop2) c2 C.df-data.frame(cbind(c0,c1,c2)) colnames(C.df)-c(c0,c1,c2) C.df # preparation of form for explaining further needed result (next 3 lines are not needed indeed, they are only to explain how to obtain final result c3-rbind(NA,Start1,Start1,Start1,Start1,Start2,Start2,Start2,Start2,Start2,Start2,Start2,Start2,Start2,Start2,Start2,Start2) c4-rbind(NA, Stop1, Stop1, Stop1, Stop1, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2, Stop2) C.df-data.frame(cbind(c0,c1,c2,c3,c4)) colnames(C.df)-c(c0,c1,c2,c3,c4) C.df$c5-paste(C.df$c3,C.df$c4,sep=-) C.df Now this is something I don't get. The list Start2-Stop2 starts way before Start2, actually at Stop1. Sure that's what you want? I took the liberty of showing how to get the data between start and stop for every entry, and how to apply functions to it. If you don't get the code, look at ?lapply ?apply ?grep I also adjusted your example, as you caused all variables to be factors by using the cbind in the data.frame function. Never do this unless you're really sure you have to. But I can't think of a case where that would be beneficial... ... C.df-data.frame(c0,c1,c2) C.df # find positions Start - grep(Start,C.df$c2) Stop - grep(Stop,C.df$c2) # create indices idx - apply(cbind(Start,Stop),1,function(i) i[1]:i[2]) names(idx) - paste(Start,1:length(Start),-Stop,1:length(Start),sep=) # Apply the function summary and get a list back named by the interval. out - lapply(idx,function(i) summary(C.df[i,1:2])) out If you really need to start Start2 right after Stop1, you can use a similar approach. Cheers Joris # NEEDED RESULTS # needed result # for Stat1-Stop1: mean(20,30,40,50) # for Stat2-Stop2: mean(c(10,60,20,30,40,50,30,10,0,NA,20,10.3444), na.rm=T) #mean: c1 c3 c4 c5 20 Start1 Stop1 Start1-Stop1 25.48585 Start2 Stop2 Start2-Stop2 #sum # for Stat1-Stop1: sum(20,30,40,50) # for Stat2-Stop2: sum(c(10,60,20,30,40,50,30,10,0,NA,20,10.3444), na.rm=T) #sum: c1 c3 c4 c5 140 Start1 Stop1 Start1-Stop1 280.3444 Start2 Stop2 Start2-Stop2 # for Stat1-Stop1: max(20,30,40,50) # for Stat2-Stop2: max(c(10,60,20,30,40,50,30,10,0,NA,20,10.3444), na.rm=T) #max: c1 c3 c4 c5 50 Start1 Stop1 Start1-Stop1 60 Start2 Stop2 Start2-Stop2 # place of max (in Start1-Stop1: 4 th element in gruop Start1-Stop1 # place of max (in Start1-Stop1: 2 nd element in
[R] GPS finding and mark interesting POI within table with large GPS points collection
Dear R users, # #There is table containing 1000 (lat, lon, time) GPS positions, wchich should be recognized and labeled in every row of that #recognized position by label from defined POI vector GPS$Lat=c(9,2.2,2,3,4,5,6,7,2,9,2.1,10,3.3,0 , 6, 8, 9) GPS$Lon=c(0,1 ,1,4,4,5,6,7,2,9,1.1,10,4.2,0 , 6, 1, 9) GPS$time =c(0,1 ,2,3,4,5,6,7,8,9,10 ,11,12 ,13,14,15,16) # and definition of points of interest: POIdef$Lat =c( 2 ,3.2) POIdef$Lon =c( 1.1 ,4 ) POIdef$Lab =c(A ,B) # but if there are two recognitized points in the small time difference, # (within time window of tw=3) # there should be choosen only one point from POIdef #Result expected: #like in these vectors: (GPSfindings$Lat =c(2.2, 2.1, 3.3) ) (GPSfindings$Lon =c(1 , 1.1, 4.2) ) (GPSfindings$time=c(2 ,10 , 12 ) ) (GPSfindings$Lab =c(A,A , B) ) (GPSfindings$POI_lat =c(2 ,2 , 3.2) ) (GPSfindings$POI_lon =c(1.1,1.1 , 4.1) ) #and as result obtain marked vector: (GPS$Lab=c(NA,A,NA,NA,NA,NA,NA,NA,NA,B,NA,NA,NA,NA,NA,NA)) #1) Is that possible without using specialized package? #2) Any specialized package function realize that? # Thanks for suggestions. Sincerely, Eugen [[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.
