Hi,
You can do this:
lst1<-lapply(split(colnames(df)[-1],gsub("_.*","",colnames(df)[-1])),function(x)
{x1<-cbind(date=df[,1],df[,x]);colnames(x1)[-1]<- x;x1})
lst1
#$ABC
# date ABC_f ABC_e ABC_d ABC_m
#1 2013-04-15 62.80740 11.36784 38.71090 40.28474
#2 2013-04-14 81.04526 62.29652 93.48216 43.97076
#3 2013-04-13 84.65712 47.63482 93.14432 47.38762
#4 2013-04-12 12.78237 32.27821 78.27387 97.33573
#5 2013-04-11 57.61345 52.12561 31.87170 22.06885
#
#$LMN
# date LMN_d LMN_a LMN_c
#1 2013-04-15 21.16794 56.67684 45.44847
#2 2013-04-14 54.65804 25.81530 17.72362
#3 2013-04-13 63.89233 40.12268 36.76901
#4 2013-04-12 87.59880 35.74175 68.58913
#5 2013-04-11 87.07694 47.95892 35.80767
#
#$PQR
# date PQR
#[1,] 5 71.22868
#[2,] 4 95.09995
#[3,] 3 83.62438
#[4,] 2 30.18525
#[5,] 1 25.81805
#
#$XYZ
# date XYZ_p XYZ_zz
#1 2013-04-15 55.88855 85.74755
#2 2013-04-14 94.13587 63.48582
#3 2013-04-13 84.00891 81.61107
#4 2013-04-12 98.99747 58.15729
#5 2013-04-11 64.71084 27.44133
lapply(seq_along(lst1),function(i)
write.csv(lst1[[i]],file=paste0(names(lst1[i]),".csv"),row.names=FALSE))
A.K.
----- Original Message -----
From: Katherine Gobin <katherine_go...@yahoo.com>
To: r-help@r-project.org
Cc:
Sent: Friday, April 26, 2013 9:21 AM
Subject: [R] Splitting data.frame and saving to csv files
Dear R Forum,
I have a data.frame as
df = data.frame(date = c("2013-04-15", "2013-04-14", "2013-04-13",
"2013-04-12", "2013-04-11"),
ABC_f = c(62.80739769,81.04525895,84.65712455,12.78237251,57.61345256),
LMN_d = c(21.16794336,54.6580401,63.8923307,87.59880367,87.07693716),
XYZ_p = c(55.8885464,94.1358684,84.0089114,98.99746696,64.71083712),
LMN_a = c(56.6768395,25.81530198,40.12268441,35.74175237,47.95892209),
ABC_e = c(11.36783959,62.29651784,47.63481552,32.27820673,52.12561419),
LMN_c = c(45.4484695,17.72362438,36.7690054,68.58912931,35.80767235),
XYZ_zz = c(85.74755089,63.48582415,81.61107212,58.1572924,27.44132817),
PQR = c(71.22867519,95.09994812,83.62437819,30.18524735,25.81804865),
ABC_d =
c(38.71089816,93.48216193,93.14432203,78.2738731,31.87170019),
ABC_m = c(40.28473769,43.97076327,47.38761559,97.33573412,22.06884976))
> df
date ABC_f LMN_d XYZ_p LMN_a ABC_e
1 2013-04-15 62.80740 21.16794 55.88855 56.67684 11.36784
2 2013-04-14 81.04526 54.65804 94.13587 25.81530 62.29652
3 2013-04-13 84.65712 63.89233 84.00891 40.12268 47.63482
4 2013-04-12 12.78237 87.59880 98.99747 35.74175 32.27821
5 2013-04-11 57.61345 87.07694 64.71084 47.95892 52.12561
LMN_c XYZ_zz PQR ABC_d ABC_m
1 45.44847 85.74755 71.22868 38.71090 40.28474
2 17.72362 63.48582 95.09995 93.48216 43.97076
3 36.76901 81.61107 83.62438 93.14432 47.38762
4 68.58913 58.15729 30.18525 78.27387
97.33573
5 35.80767 27.44133 25.81805 31.87170 22.06885
I need to identify columns with same labels and along-with the dates in the
first column, save the columns in different csv files.
E.g. in the above data frame, I have 4 columns beginning with ABC so I need to
save these four columns with the date in the first column as ABC.csv, then
LMN_d, LMN_a, LMN_c in the LMN.csv file as date, LMN_a, LMN_c, LMN_d and so on.
In my actual data.frame, I won't be aware how many such rates combinations are
available. If there is no matching column as "PQR", the PQR.csv file should
have only date and PQR column.
Kindly guide how do I split the data.frame and save the respective csv files.
Regards
Katherine
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______________________________________________
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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.
