[R] How to store interim print results
Dear R forum,
Following is an customized extract of a code I am working on.
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
{
nC - seq(date, by=paste (n, months), length = 2)[2]
fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
if(nCC) return(C)
return(nC)
}
date.diff = function(end, start, basis=1) {
if (basis != 0 basis != 4)
return(as.numeric(end - start))
e - as.POSIXlt(end)
s - as.POSIXlt(start)
d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
return (d)
}
cashflows - 0
last.coupon - maturity
while (last.coupon settlement) {
print(last.coupon) # I need to store these dates
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows) # I need to store these cashflow numbers
}
The print command causes the following output
[1] 2015-10-01
[1] 1
[1] 2015-04-01
[1] 2
[1] 2014-10-01
[1] 3
[1] 2014-04-01
[1] 4
My problem is how do I store these print outputs or while the loop is getting
executed, how do I save these to some data.frame say
output_dat
cashflow_tenure cashflow_nos
1 2015-10-01 1
2 2015-04-01 2
3 2014-10-01 3
4 2014-04-01 4
Kindly advise
With regards
Katherine
[[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 store interim print results
This will get you close:
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
+ {
+ nC - seq(date, by=paste (n, months), length = 2)[2]
+ fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
+ C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
+ if(nCC) return(C)
+ return(nC)
+ }
date.diff = function(end, start, basis=1) {
+ if (basis != 0 basis != 4)
+ return(as.numeric(end - start))
+ e - as.POSIXlt(end)
+ s - as.POSIXlt(start)
+ d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
+
+ return (d)
+ }
output - capture.output({ # collect the print output
+ cashflows - 0
+ last.coupon - maturity
+ while (last.coupon settlement) {
+ print(last.coupon) # I need to store these dates
+ last.coupon - add.months(last.coupon, -12/frequency)
+ cashflows - cashflows + 1
+ print(cashflows) # I need to store these cashflow numbers
+ }
+
+ })
# remove line numbers
output - sub(^, , output)
# remove extra quotes
output - gsub('', '', output)
# now read in the data
report - matrix(output, ncol = 2, byrow = TRUE)
report
[,1] [,2]
[1,] 2015-10-01 1
[2,] 2015-04-01 2
[3,] 2014-10-01 3
[4,] 2014-04-01 4
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Apr 3, 2014 at 6:22 AM, Katherine Gobin
katherine_go...@yahoo.comwrote:
Dear R forum,
Following is an customized extract of a code I am working on.
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
{
nC - seq(date, by=paste (n, months), length = 2)[2]
fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
if(nCC) return(C)
return(nC)
}
date.diff = function(end, start, basis=1) {
if (basis != 0 basis != 4)
return(as.numeric(end - start))
e - as.POSIXlt(end)
s - as.POSIXlt(start)
d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
return (d)
}
cashflows - 0
last.coupon - maturity
while (last.coupon settlement) {
print(last.coupon) # I need to store these dates
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows) # I need to store these cashflow numbers
}
The print command causes the following output
[1] 2015-10-01
[1] 1
[1] 2015-04-01
[1] 2
[1] 2014-10-01
[1] 3
[1] 2014-04-01
[1] 4
My problem is how do I store these print outputs or while the loop is
getting executed, how do I save these to some data.frame say
output_dat
cashflow_tenurecashflow_nos
1 2015-10-011
2 2015-04-012
3 2014-10-013
4 2014-04-014
Kindly advise
With regards
Katherine
[[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.
[[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 store interim print results
Katherine,
One easy way to do this for small data is by using the append() function
(see code below). But, if you have a lot of data, it may be too slow for
you. In that case, you can gain some efficiency if you determine in
advance how long the vectors will be, then use indexing to fill in the
vectors without using the append() function. Or, rewrite the code to be
vectorized instead of using a while() loop.
cashflows - 0
last.coupon - maturity
# create empty vectors
cashflow.tenure - character(0)
cashflow.nos - numeric(0)
while (last.coupon settlement) {
print(last.coupon)
# store the dates
cashflow.tenure - append(cashflow.tenure, last.coupon)
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows)
# store the cashflow numbers
cashflow.nos - append(cashflow.nos, cashflows)
}
output.dat - data.frame(cashflow.tenure, cashflow.nos)
output.dat
Jean
On Thu, Apr 3, 2014 at 5:22 AM, Katherine Gobin
katherine_go...@yahoo.comwrote:
Dear R forum,
Following is an customized extract of a code I am working on.
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
{
nC - seq(date, by=paste (n, months), length = 2)[2]
fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
if(nCC) return(C)
return(nC)
}
date.diff = function(end, start, basis=1) {
if (basis != 0 basis != 4)
return(as.numeric(end - start))
e - as.POSIXlt(end)
s - as.POSIXlt(start)
d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
return (d)
}
cashflows - 0
last.coupon - maturity
while (last.coupon settlement) {
print(last.coupon) # I need to store these dates
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows) # I need to store these cashflow numbers
}
The print command causes the following output
[1] 2015-10-01
[1] 1
[1] 2015-04-01
[1] 2
[1] 2014-10-01
[1] 3
[1] 2014-04-01
[1] 4
My problem is how do I store these print outputs or while the loop is
getting executed, how do I save these to some data.frame say
output_dat
cashflow_tenurecashflow_nos
1 2015-10-011
2 2015-04-012
3 2014-10-013
4 2014-04-014
Kindly advise
With regards
Katherine
[[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.
[[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 store interim print results
Dear Jean
Thanks a lot for this solution. Its very useful. I did only one small change
and defined
cashflow.tenure - numeric(0) instead of character(0). This helps me in further
numerical calculations using these dates like finding the difference between
two dates etc.
Thanks again,
Regards
Katherine
On Thursday, 3 April 2014 6:58 PM, Adams, Jean jvad...@usgs.gov wrote:
Katherine,
One easy way to do this for small data is by using the append() function (see
code below). But, if you have a lot of data, it may be too slow for you. In
that case, you can gain some efficiency if you determine in advance how long
the vectors will be, then use indexing to fill in the vectors without using the
append() function. Or, rewrite the code to be vectorized instead of using a
while() loop.
cashflows - 0
last.coupon - maturity
# create empty vectors
cashflow.tenure - character(0)
cashflow.nos - numeric(0)
while (last.coupon settlement) {
print(last.coupon)
# store the dates
cashflow.tenure - append(cashflow.tenure, last.coupon)
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows)
# store the cashflow numbers
cashflow.nos - append(cashflow.nos, cashflows)
}
output.dat - data.frame(cashflow.tenure, cashflow.nos)
output.dat
Jean
On Thu, Apr 3, 2014 at 5:22 AM, Katherine Gobin katherine_go...@yahoo.com
wrote:
Dear R forum,
Following is an customized extract of a code I am working on.
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
{
nC - seq(date, by=paste (n, months), length = 2)[2]
fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
if(nCC) return(C)
return(nC)
}
date.diff = function(end, start, basis=1) {
if (basis != 0 basis != 4)
return(as.numeric(end - start))
e - as.POSIXlt(end)
s - as.POSIXlt(start)
d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
return (d)
}
cashflows - 0
last.coupon - maturity
while (last.coupon settlement) {
print(last.coupon) # I need to store these dates
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows) # I need to store these cashflow numbers
}
The print command causes the following output
[1] 2015-10-01
[1] 1
[1] 2015-04-01
[1] 2
[1] 2014-10-01
[1] 3
[1] 2014-04-01
[1] 4
My problem is how do I store these print outputs or while the loop is getting
executed, how do I save these to some data.frame say
output_dat
cashflow_tenure cashflow_nos
1 2015-10-01 1
2 2015-04-01 2
3 2014-10-01 3
4 2014-04-01 4
Kindly advise
With regards
Katherine
[[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.
[[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 store interim print results
Dear Sir,
Thanks a lot for your guidance and efforts. Appreciate it.
Thanks again.
Katherine
On Thursday, 3 April 2014 6:55 PM, jim holtman jholt...@gmail.com wrote:
This will get you close:
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
+ {
+ nC - seq(date, by=paste (n, months), length = 2)[2]
+ fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
+ C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
+ if(nCC) return(C)
+ return(nC)
+ }
date.diff = function(end, start, basis=1) {
+ if (basis != 0 basis != 4)
+ return(as.numeric(end - start))
+ e - as.POSIXlt(end)
+ s - as.POSIXlt(start)
+ d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
+
+ return (d)
+ }
output - capture.output({ # collect the print output
+ cashflows - 0
+ last.coupon - maturity
+ while (last.coupon settlement) {
+ print(last.coupon) # I need to store these dates
+ last.coupon - add.months(last.coupon, -12/frequency)
+ cashflows - cashflows + 1
+ print(cashflows) # I need to store these cashflow numbers
+ }
+
+ })
# remove line numbers
output - sub(^, , output)
# remove extra quotes
output - gsub('', '', output)
# now read in the data
report - matrix(output, ncol = 2, byrow = TRUE)
report
[,1] [,2]
[1,] 2015-10-01 1
[2,] 2015-04-01 2
[3,] 2014-10-01 3
[4,] 2014-04-01 4
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Apr 3, 2014 at 6:22 AM, Katherine Gobin katherine_go...@yahoo.com
wrote:
Dear R forum,
Following is an customized extract of a code I am working on.
settlement = as.Date(2013-11-25)
maturity = as.Date(2015-10-01)
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
{
nC - seq(date, by=paste (n, months), length = 2)[2]
fD - as.Date(strftime(as.Date(date), format='%Y-%m-01'))
C - (seq(fD, by=paste (n+1, months), length = 2)[2])-1
if(nCC) return(C)
return(nC)
}
date.diff = function(end, start, basis=1) {
if (basis != 0 basis != 4)
return(as.numeric(end - start))
e - as.POSIXlt(end)
s - as.POSIXlt(start)
d - (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
return (d)
}
cashflows - 0
last.coupon - maturity
while (last.coupon settlement) {
print(last.coupon) # I need to store these dates
last.coupon - add.months(last.coupon, -12/frequency)
cashflows - cashflows + 1
print(cashflows) # I need to store these cashflow numbers
}
The print command causes the following output
[1] 2015-10-01
[1] 1
[1] 2015-04-01
[1] 2
[1] 2014-10-01
[1] 3
[1] 2014-04-01
[1] 4
My problem is how do I store these print outputs or while the loop is getting
executed, how do I save these to some data.frame say
output_dat
cashflow_tenure cashflow_nos
1 2015-10-01 1
2 2015-04-01 2
3 2014-10-01 3
4 2014-04-01 4
Kindly advise
With regards
Katherine
[[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.
[[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.
