Thank you so much.
You are the BEST:-)
Now, I'm gonna learn all these codes and understand.
________________________________
Cc: R help <r-help@r-project.org>
Sent: Saturday, April 13, 2013 5:18 PM
Subject: Re: Comparison of Date format
Hi,
DataA<- structure(list(ID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 3L, 3L, 3L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L,
8L, 8L, 8L, 8L), Status = c("A", "B", "A", "B", "A", "B", "A",
"A", "B", "B", "A", "A", "A", "A", "B", "A", "B", "A", "A", "A",
"B", "A", "B", "A", "B", "A", "B", "A", "A", "B", "A", "B", "A",
"B", "A", "B", "A", "A", "B", "A", "B", "A", "B", "A", "B", "A"
), Date1 = c("03-Feb-01", "15-May-01", "16-May-01", "06-Sep-01",
"13-Sep-01", "26-Feb-04", "08-Mar-04", "09-Feb-01", "24-May-01",
"25-May-01", "06-Mar-02", "18-Mar-02", "20-Mar-02", "01-Apr-01",
"05-May-01", "08-Jul-01", "09-Feb-01", "13-Jun-01", "14-Jun-01",
"19-Jun-01", "05-Jul-01", "29-Oct-01", "13-Nov-01", "19-Jan-02",
"05-Feb-02", "18-Mar-02", "01-Apr-02", "30-Oct-02", "21-Nov-02",
"26-Mar-03", "06-Jun-03", "17-Jun-03", "05-Nov-03", "12-Nov-03",
"13-Nov-03", "01-Apr-02", "30-Oct-02", "21-Nov-02", "26-Mar-03",
"06-Jun-03", "17-Jun-03", "05-Nov-03", "12-Nov-03", "13-Nov-03",
"12-Nov-03", "13-Nov-03"), Date2 = c("15-May-01", "16-May-01",
"06-Sep-01", "13-Sep-01", "26-Feb-04", "08-Mar-04", "01-Aug-04",
"24-May-01", "25-May-01", "06-Mar-02", "18-Mar-02", "20-Mar-02",
"27-Mar-02", "08-May-02", "14-Jun-02", "07-Sep-02", "13-Jun-01",
"14-Jun-01", "19-Jun-01", "05-Jul-01", "29-Oct-01", "13-Nov-01",
"19-Jan-02", "05-Feb-02", "18-Mar-02", "01-Apr-02", "30-Oct-02",
"21-Nov-02", "26-Mar-03", "06-Jun-03", "17-Jun-03", "05-Nov-03",
"12-Nov-03", "13-Nov-03", "21-Nov-03", "30-Oct-02", "21-Nov-02",
"26-Mar-03", "06-Jun-03", "17-Jun-03", "05-Nov-03", "12-Nov-03",
"13-Nov-03", "21-Nov-03", "13-Nov-03", "21-Nov-03")), .Names = c("ID",
"Status", "Date1", "Date2"), class = "data.frame", row.names = c(NA,
-46L))
DataB<- structure(list(ID = c(1L, 1L, 2L, 8L, 8L, 8L, 8L, 8L), Date.Accident =
c("06-Sep-01",
"20-Jan-05", "06-Mar-02", "25-Jun-01", "29-Oct-01", "18-Mar-02",
"06-Jun-03", "16-Jan-04")), .Names = c("ID", "Date.Accident"), row.names =
c(NA,
-8L), class = "data.frame")
library(stringr)
DataB[,2]<- str_trim(DataB[,2])
DataA<- read.csv("FarnooshDataA.csv",sep=",",stringsAsFactors=FALSE)
DataA[,3]<- str_trim(DataA[,3])
DataA[,4]<- str_trim(DataA[,4])
dataNew1<-do.call(rbind,lapply(unique(DataA$ID),function(i) {x1<-
DataA[DataA$ID==i,]; x2<-DataB[DataB$ID==i,];
do.call(rbind,lapply(seq_len(nrow(x2)),function(i){x3<-
unlist(mapply(function(x,y) which(x%in% y),x1[,3:4],x2[i,2]));
x3<-if(is.matrix(x3)) x3[1,] else x3; x4<- if(length(x3)==2)
x1[x3[which.min(x3)],!names(x1)%in% names(x3[which.max(x3)])] else
if(length(x3)==1) x1[x3,c("ID","Status",names(x3))] else NULL;
if(length(x4)>0){colnames(x4)[3]<- colnames(DataB)[2];x4} else NULL})) }))
DataA1<- DataA
DataA1$Date1<-as.Date(DataA1$Date1,format="%d-%b-%y")
DataA1$Date2<-as.Date(DataA1$Date2,format="%d-%b-%y")
DataB1<- DataB
DataB1$Date.Accident<- as.Date(DataB1$Date.Accident,format="%d-%b-%y")
dataNew2<-do.call(rbind,lapply(unique(DataA1$ID),function(i) { x1<-
DataA1[DataA1$ID==i,] ;x1New<- t(x1); x2<-DataB1[DataB1$ID==i,];
do.call(rbind,lapply(seq_len(nrow(x2)),function(i){x3<-x1[x2[i,2]>x1New[3,] &
x2[i,2]< x1New[4,] ,]; x4<-x2[i,][nrow(x3)>0];merge(x3[,1:2],x4) }))}))
dataNew2$Date.Accident<-format(dataNew2$Date.Accident,"%d-%b-%y")
library(plyr)
res<-join(rbind(dataNew1,dataNew2),DataB,by=c("ID","Date.Accident"),type="right")
res
# ID Date.Accident Status
#1 1 06-Sep-01 A
#2 1 20-Jan-05 <NA>
#3 2 06-Mar-02 B
#4 8 25-Jun-01 A
#5 8 29-Oct-01 B
#6 8 18-Mar-02 B
#7 8 06-Jun-03 B
#8 8 16-Jan-04 <NA>
dat2<- data.frame(ID=unique(DataA$ID))
res2<-join(dat2,res,by="ID",type="full")
res2$Status[is.na(res2$Date.Accident) & is.na(res2$Status)]<-0
res2
# ID Date.Accident Status
#1 1 06-Sep-01 A
#2 1 20-Jan-05 <NA>
#3 2 06-Mar-02 B
#4 3 <NA> 0
#5 8 25-Jun-01 A
#6 8 29-Oct-01 B
#7 8 18-Mar-02 B
#8 8 06-Jun-03 B
#9 8 16-Jan-04 <NA>
This is similar to the output you wanted in the attached excel sheet except
that the status for "29-Oct-01" should be B as per the rules.
A.K.
ID Date.Accident Result
06-Sep-01 A
1 20-Jan-05 NA
2 06-Mar-02 B
25-Jun-01 A
29-Oct-01 A
8 18-Mar-02 B
06-Jun-03 B
16-Jan-04 NA
________________________________
Sent: Saturday, April 13, 2013 3:39 PM
Subject: Re: Comparison of Date format
Hi Arun,
Thanks you so much for your help and effort.
I have attached an excel file that I put the conditions in it. It might help
for coding.
I basically want to check if the dateAccident is between date1 and
date2--->return(status in that row)
If dateAccident =date1 and date2 ---> return (Status for Previous row)
If dateAccident=date1 and itsNOT equal to date2 --->return the status for date1
If ID in DateA not found in ID in DateB---> (I'm not sure what I can return
here) I can return only ID with status 0(?)
Thanks again.
Best,Farnoosh Sheikhi
________________________________
Cc: R help <r-help@r-project.org>
Sent: Friday, April 12, 2013 10:19 PM
Subject: Re: Comparison of Date format
HI,
In cases like below:
DataA<- read.table(text="
ID,Status,Date1,Date2
1,A,3-Feb-01,15-May-01
1,B,15-May-01,16-May-01
1,A,16-May-01,3-Sep-01
1,B,3-Sep-01,13-Sep-01
1,C,13-Sep-01,26-Feb-04
2,A,9-Feb-01,24-May-01
2,B,24-May-01,25-May-01
2,A,25-May-01,16-Mar-02
2,A,6-Mar-02,18-Mar-02
2,A,14-Sep-01,6-Mar-02
",sep=",",header=TRUE,stringsAsFactors=FALSE)
library(stringr)
DataA[,3]<- str_trim(DataA[,3])
DataA[,4]<- str_trim(DataA[,4])
DataB<- read.table(text="
ID Date.Accident
1 3-Sep-01
1 20-Jan-05
1 26-Feb-04
2 25-May-01
2 6-Mar-02
",sep="",header=TRUE,stringsAsFactors=FALSE)
DataB[,2]<- str_trim(DataB[,2])
#Needed modification in the 2nd method:
dataNew2<-do.call(rbind,lapply(unique(DataA$ID),function(i){x1<-
DataA[DataA$ID==i,]; x2<- DataB[DataB$ID==i,];
do.call(rbind,lapply(seq_len(nrow(x2)),function(i) {x3<-
unlist(mapply(function(x,y) which(x==y), x1[,3:4],x2[i,2])); x4<-
if(length(x3)==2) x1[x3[which.min(x3)],!names(x1)%in%names(x3[which.max(x3)])]
else if(length(x3)==1) x1[x3,c("ID","Status",names(x3))] else NULL;
if(length(x4)>0) {colnames(x4)[3]<- colnames(DataB)[2]; x4} else NULL})) }))
library(plyr)
res2<- join(dataNew2,DataB,by=c("Date.Accident","ID"),type="right")
res2
# Date.Accident ID Status
#1 3-Sep-01 1 A
#2 20-Jan-05 1 <NA>
#3 26-Feb-04 1 C
#4 25-May-01 2 B
#5 6-Mar-02 2 A
Using the first
method:
lst1<-lapply(seq_len(nrow(DataB)),function(i) {x1<-unlist(mapply(function(x,y)
which(x==y),DataA[,3:4],DataB[i,2]));x2<-if(length(x1)==2)
DataA[x1[which.min(x1)],!names(DataA)%in%names(x1[which.max(x1)])] else
if(length(x1)==1) DataA[x1,c("ID","Status",names(x1))] else NULL})
lst2<-lapply(lst1,data.frame)
lst2<-lst2[lapply(lst2,nrow)!=0]
library(plyr)
dataNew<-do.call(rbind,lapply(lst2,function(x) {colnames(x)[3]<-
colnames(DataB)[2];x}))
res<-join(dataNew,DataB,by=c("Date.Accident","ID"),type="right")
identical(res,res2)
#[1] TRUE
A.K.
----- Original Message -----
Cc: R help <r-help@r-project.org>
Sent: Saturday, April 13, 2013 12:41 AM
Subject: Re: Comparison of Date format
Hi,
In the example you provided, it looks like the dates in Date2 happens first.
So, I changed it a bit.
DataA<- read.table(text="
ID,Status,Date1,Date2
1,A,3-Feb-01,15-May-01
1,B,15-May-01,16-May-01
1,A,16-May-01,3-Sep-01
1,B,3-Sep-01,13-Sep-01
1,C,13-Sep-01,26-Feb-04
2,A,9-Feb-01,24-May-01
2,B,24-May-01,25-May-01
2,A,25-May-01,16-Mar-02
2,A,6-Mar-02,18-Mar-02
2,A,14-Sep-01,6-Mar-02
",sep=",",header=TRUE,stringsAsFactors=FALSE)
library(stringr)
DataA[,3]<- str_trim(DataA[,3])
DataA[,4]<- str_trim(DataA[,4])
DataB<- read.table(text="
ID Date.Accident
1 3-Sep-01
1 20-Jan-05
1 26-Feb-04
2
6-Mar-02
",sep="",header=TRUE,stringsAsFactors=FALSE)
lst1<-lapply(seq_len(nrow(DataB)),function(i) {x1<-unlist(mapply(function(x,y)
which(x==y),DataA[,3:4],DataB[i,2]));x2<-if(length(x1)==2)
DataA[x1[which.min(x1)],!names(DataA)%in%names(x1[which.max(x1)])] else
if(length(x1)==1) DataA[x1,c("ID","Status",names(x1))] else NULL})
lst2<-lapply(lst1,data.frame)
lst2<-lst2[lapply(lst2,nrow)!=0]
lst2
#[[1]]
# ID Status Date2
#3 1 A 3-Sep-01
#[[2]]
# ID Status Date2
#5 1 C 26-Feb-04
#[[3]]
# ID Status Date1
#9 2 A 6-Mar-02
library(plyr)
dataNew<-do.call(rbind,lapply(lst2,function(x) {colnames(x)[3]<-
colnames(DataB)[2];x}))
res<-join(dataNew,DataB,by=c("Date.Accident","ID"),type="right")
res
# Date.Accident ID Status
#1 3-Sep-01 1 A
#2 20-Jan-05 1 <NA>
#3 26-Feb-04 1 C
#4 6-Mar-02 2 A
#or you can split by ID
lst1New<-lapply(unique(DataA$ID),function(i){x1<- DataA[DataA$ID==i,]; x2<-
DataB[DataB$ID==i,]; do.call(rbind,lapply(seq_len(nrow(x2)),function(i) {x3<-
unlist(mapply(function(x,y) which(x==y), x1[,3:4],x2[i,2])); x4<-
if(length(x3)==2) x1[x3[which.min(x3)],!names(x1)%in%names(x3[which.max(x3)])]
else if(length(x3)==1) x1[x3,c("ID","Status",names(x3))] else NULL})) })
lst1New
#[[1]]
# ID
Status Date2
#3 1 A 3-Sep-01
#5 1 C 26-Feb-04
#[[2]]
# ID Status Date1
#9 2 A 6-Mar-02
dataNew1<- do.call(rbind,lapply(lst1New,function(x) {colnames(x)[3]<-
colnames(DataB)[2];x}))
res1<- join(dataNew1,DataB,by=c("Date.Accident","ID"),type="right")
res1
# Date.Accident ID Status
#1 3-Sep-01 1 A
#2 20-Jan-05 1 <NA>
#3 26-Feb-04 1 C
#4 6-Mar-02 2 A
A.K.
________________________________
Sent: Friday, April 12, 2013 5:40 PM
Subject: Comparison of Date format
Hi there,
Hope all is well.
I have a complicated data and I need to create a new variable based on the date
and description of the data.
I really appreciate if you can help me.
Here is how data look like:
DataA
DataB
ID Status Date1 Date2
ID Date.Accident
Result
1 A 3-Feb-01 15-May-01
1 3-Sep-01
A
1 B 15-May-01 16-May-01
1 20-Jan-05
NA
1 A 16-May-01 3-Sep-01
1
B 3-Sep-01 13-Sep-01
1 C 13-Sep-01 26-Feb-04
2 A 9-Feb-01 24-May-01
2 6-Mar-02
A
2 B 24-May-01 25-May-01
2 A 25-May-01 6-Mar-02
2 A 6-Mar-02 18-Mar-02
I want to compare dataA to B for each ID. if Date 1 or Date 2 matches to
Date.Accident return the result as status in dataA as a new result in Data B.
The trick here is I have two dates that are matched, but I want the status of
the one that happen first. The sample size of each data is not the same.
I really appreciate your time and help.
Thanks.
[[alternative HTML version deleted]]
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