[R] Reading multiple csv files
Dear R helpers
Some particular analysis leads me to various number of output csv files
depending on some conditions. Say e.g. I have output files variable1.csv,
variable2.csv, .. Problem is I don't know how many csv files been
generated. They could be 4, 5 or even 10. Each file will have a column called
amount.
My problem is to find filewise mean(amount) and sd(amount). I need to write a
loop where all these individual csv files will be read and after reading each
file, mean and sd will be calculated.
I have tried to write some R code which is very absurd.
for (i in 1 : n) # n is no of input files
{
data[i] = read.csv(file = paste("variable", i, ".csv", sep = ""))$amount
mean(data[i])
sd(data[i])
}
I get following error.
Error in file(file, "rt") : cannot open the connection
In addition: Warning message:
In file(file, "rt") :
cannot open file 'paste("output", i, ".csv", sep = "")': Invalid argument
Please guide
Regards
Madhavi
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[R] How to use same function for diffrent input values
Dear R helpers
I have written some function (the actual code I have pasted at the end of mail)
like say
indiv_rate = function(n, rate_name, rate, rate_rf1, rate_rf2, rate_rf3,
rateprob1, rateprob2, rateprob3)
{
some R commands
return(data.frame(rate_name, rates = round(rate_data, digits = 4)))
}
## INPUT
rates = indiv_rate(n = read.csv('number.csv')$n, rate_name =
read.csv('rate.csv')$rate_name, rate = read.csv('rate.csv')$rate, rate_rf1 =
read.csv('rate_rf.csv')$rate_rf1,
rate_rf2 = read.csv('rate_rf.csv')$rate_rf2, rate_rf3
= read.csv('rate_rf.csv')$rate_rf3,
rateprob1 = read.csv('rate_probability.csv')$probability1, rateprob2 =
read.csv('rate_probability.csv')$probability2, rateprob3 =
read.csv('rate_probability.csv')$probability3)
## OUTPUT
write.csv(data.frame(rates), 'indiv rates generated.csv', row.names = FALSE)
##___ end of code
So for given rate (which I am deining as rate = read.csv('rate.csv')$rate), I
get the desired results.
My problem is how do I use this fuction for different 'rates' i.e. for any
given rate, I run the function and store the result with the respective rate
name?
Regards
Madhavi
(PS - I am avoiding to paste my actual code consuming 60 lines. Still, if
somene insists, I can post the same)
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Re: [R] Total and heading of portfoilo table
Hi!
I am not expert in R, but perhaps you can try the following -
X = as.numeric(read.csv('quantity.csv'))
Y = read.csv('equity_price.csv')
Y = Y[, -1]
Z = X*Y
port_val = NULL
for(i in 1 : nrow(Z))
{
port_val[i] = sum(Z[i,])
}
write.csv(data.frame(Z, port_val = port_val), 'PORTFOLIO.csv', row.names =
FALSE)
I am sure the experts will have much simpler way to address this problem.
Regards
Madhavi
--- On Mon, 15/2/10, Sarah Sanchez wrote:
From: Sarah Sanchez
Subject: [R] Total and heading of portfoilo table
To: r-help@r-project.org
Date: Monday, 15 February, 2010, 10:08 PM
Dear R helpers,
I have two input files as 'quantity.csv' and 'equity_price.csv' as (for
example) given below.
'quantity.csv'
GOOG YHOO
1000 100
'equity_price.csv'
sr_no GOOG_price YHOO_price
1 15.22 536.40
2 15.07 532.97
3 15.19 534.05
4 15.16 531.86
5 15.11 532.11
My problem is to calculate the portfolio value for each of these 5 days
(actually my portfolio
consists of 47 comanies and prices taken are for last 1 year).
I had defined
X = read.csv('quantity.csv')
Y = read.csv('equity_price.csv')
I have tried the loop
Z = array()
for (i in 1:2)
{
Z[i] = (X[[i]]*Y[i])
}
# When I write this dataframe as
write.csv(data.frame(Z), 'Z.csv', row.names = FALSE)
When I open 'Z.csv' file, I get
c.2500L..3300L..4500L..1000L..4400L.
c.14000L..45000L..48000L..26000L..15000L.
2500 14000
3300 45000
4500 48000
1000 26000
4400 15000
My requirement is to have the column heads and the portfolio total as
GOOG YHOO Total
2500 14000 16500
3300 45000 48300
4500 48000 52500
1000 26000 27000
4400 15000 19400
Please guide
Regards
Sarah
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Re: [R] CORRECTION - Storing results in a loop
Dear Sir,
Thanks a lot for your quick solution. But still I will like someone to guide me
the solution as per my requirement. The problem is actually I am not looking
for the square of each term. I have used it to give some example since I didn't
want to confuse the matter. My process is altogether different and I need to
use loop. So here is my actual code. I am trying to use Historical simulation
and for that I need to calculate LN(New rate / old rate for each of the
instrument separately).
Here is my actual code -
ONS = read.csv('Instrument.csv')
n = length(ONS)
Y = NULL
B = array()
for (i in 1 : n)
{
Y[i] = ONS[i]
j <- 1
for (j in 1:(length(Y[[i]])-1))
{
B[j] <- log((Y[[i]][j+1])/(Y[[i]][j]))
j <- j+1
}
}
--- On Mon, 15/2/10, Benilton Carvalho wrote:
From: Benilton Carvalho
Subject: Re: [R] CORRECTION - Storing results in a loop
To: "Madhavi Bhave"
Cc: r-help@r-project.org
Date: Monday, 15 February, 2010, 4:29 AM
sorry, meant to type:
B = ONS^2
cheers,
benilton
On Mon, Feb 15, 2010 at 12:28 PM, Benilton Carvalho
wrote:
> maybe you just want
>
> Y = ONS^2
>
> ?
>
> b
>
> On Mon, Feb 15, 2010 at 12:22 PM, Madhavi Bhave
> wrote:
>> Dear R Helpers
>>
>> (There is a small correction in my earlier mail. In the 'instrument.csv'
>> file, I had mentioned only three columns. Actually there are 7 columns. I
>> regret the error. Rest contents remains the same. Thanks)
>>
>> I have an 'instrument.csv' file with 7 instrument names and 5 rates each
>> i.e. it has 7 columns and 6 rows (including row names).
>>
>> 'instrument.csv'
>>
>> instrument1 instrument2 instrument7
>> 12 5 14
>> 11 7 7
>> 14 11 3
>> 8 21 10
>> 11 3 5
>>
>>
>> Following is my R code.
>>
>> ONS = read.csv('Instrument.csv')
>> n = length(ONS)
>>
>> Y = NULL
>> B = NULL
>>
>> for (i in 1 : n)
>>
>> {
>>
>> Y[i] = ONS[i]
>>
>> for (j in 1 : length(Y[[i]]))
>> {
>> B[j] = (Y[[i]][j])^2
>> }
>>
>> }
>>
>> Problem is when I type B, I get the processed result only for the last
>> column i.e. Y[7]. It doesn't store results for Y[1] to Y[7].
>>
>> I need B[1], B[2]...upto B[7].
>>
>> Please guide me how do I store individual column processed results?
>>
>> Thanking you all in advance
>>
>> Regards
>>
>> Madhavi
>>
>>
>>
>>
>> The INTERNET now has a personality. YOURS! See your Yahoo! Homepage.
>> [[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.
>>
>>
>
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[R] CORRECTION - Storing results in a loop
Dear R Helpers
(There is a small correction in my earlier mail. In the 'instrument.csv' file,
I had mentioned only three columns. Actually there are 7 columns. I regret the
error. Rest contents remains the same. Thanks)
I have an 'instrument.csv' file with 7 instrument names and 5 rates each i.e.
it has 7 columns and 6 rows (including row names).
'instrument.csv'
instrument1 instrument2 instrument7
12 5 14
11 7 7
14 11 3
8 21 10
11 3 5
Following is my R code.
ONS = read.csv('Instrument.csv')
n = length(ONS)
Y = NULL
B = NULL
for (i in 1 : n)
{
Y[i] = ONS[i]
for (j in 1 : length(Y[[i]]))
{
B[j] = (Y[[i]][j])^2
}
}
Problem is when I type B, I get the processed result only for the last column
i.e. Y[7]. It doesn't store results for Y[1] to Y[7].
I need B[1], B[2]...upto B[7].
Please guide me how do I store individual column processed results?
Thanking you all in advance
Regards
Madhavi
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[R] Storing processed results in a loop
Dear R Helpers
I have an 'instrument.csv' file with 3 instrument names and 5 rates each i.e.
it has 7 columns and 6 rows (including row names).
'instrument.csv'
instrument1 instrument2 instrument3
12 5 14
11 7 7
14 11 3
8 2110
11 3 5
Following is my R code.
ONS = read.csv('Instrument.csv')
n = length(ONS)
Y = NULL
B = NULL
for (i in 1 : n)
{
Y[i] = ONS[i]
for (j in 1 : length(Y[[i]]))
{
B[j] = (Y[[i]][j])^2
}
}
Problem is when I type B, I get the processed result only for the last column
i.e. Y[7]. It doesn't store results for Y[1] to Y[7].
I need B[1], B[2]...upto B[7].
Please guide me how do I store individual column processed results?
Thanking you all in advance
Regards
Madhavi
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Re: [R] How to repeat the names?
Dear Sirs,
Thanks a lot for your guidance. It worked wonderfully.
Regards
Madhavi
--- On Wed, 10/2/10, Ivan Calandra wrote:
From: Ivan Calandra
Subject: Re: [R] How to repeat the names?
To: r-help@r-project.org
Date: Wednesday, 10 February, 2010, 1:54 AM
Hi!
I'm kind of a newboe here, but I think it is because read.csv transforms
the character variables to factors. Maybe try setting the argument
"as.is" in read.csv().
Or try:
rep(as.character(c(city1, city2)),5)
That should work.
HTH
Ivan
Le 2/10/2010 10:44, Madhavi Bhave a écrit :
> Dear R helpers
>
> I have a city.csv file as given below.
>
> 'city.csv'
> city_name1 city_name2
> New York City Buffallo
>
> So I define
>
> city_name = read.csv('city.csv')
> city1 = city_name$city_name1
> city2 = city_name$city_name2
>
> My problem is how do I repeat the names one after other say 10 times i.e. my
> output should be like
>
> New York
> City Buffallo
> New York
> City Buffallo
> New York
> City Buffallo
> New York City
> ...
> ...
> ...
> ...
>
> I have tried the following commands
>
> rep(c(city1,city2), 5)
>
> and I got the output something like this
>
> [1] 1 1 1 1 1 1 ...
>
> If I try
>
> rep((city1,city2), 5)
>
> Error: unexpected ',' in "rep((city1,"
>
> Please guide
>
> Regards
>
> MAdhavi
>
>
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>
>
>
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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[R] How to repeat the names?
Dear R helpers
I have a city.csv file as given below.
'city.csv'
city_name1 city_name2
New York City Buffallo
So I define
city_name = read.csv('city.csv')
city1 = city_name$city_name1
city2 = city_name$city_name2
My problem is how do I repeat the names one after other say 10 times i.e. my
output should be like
New York
City Buffallo
New York
City Buffallo
New York
City Buffallo
New York City
...
...
...
...
I have tried the following commands
rep(c(city1,city2), 5)
and I got the output something like this
[1] 1 1 1 1 1 1 ...
If I try
rep((city1,city2), 5)
Error: unexpected ',' in "rep((city1,"
Please guide
Regards
MAdhavi
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Re: [R] Yield to Maturity using R
Dear Sir,
Thank you for valuable guidance. Though I have been using R occassionally, it
was limited to some basics and that way I am new to R. As suggested by you, I
have gone through the said chapter of Introduction to R manual, though I have
some urgent comittments to meet.
I have tried writing function as given below.
f = function(price, tenure, no_comp, coupon_rate, face_value)
{
coupon_payment = face_value * coupon_rate / no_comp
cash_flow = c(rep(c(coupon_payment), (no_comp * tenure - 1)), face_value +
coupon_payment)
E = NULL
for (i in 1 : (tenure * no_comp - 1))
{
E[i] = cash_flow[i]/(1+ytm)^i
}
F = NULL
{
F = sum(E) + ((face_value + coupon_payment)/(1+ytm)^(no_comp * tenure)) -
price
}
return(data.frame(S = uniroot.all(F, interval=c(0,25
}
output = f(1010, 3, 1, 0.10, 1000)
## End of code
However, when I try to execute the same, I get following error.
Error: object 'ytm' not found
My objective is to find ytm itself and I am not able to figure out where I am
going wrong and how to overcome the same.
Regards
Madhavi
--- On Tue, 2/2/10, Dennis Murphy wrote:
From: Dennis Murphy
Subject: Re: [R] Yield to Maturity using R
To: "Madhavi Bhave"
Date: Tuesday, 2 February, 2010, 3:49 AM
Hi:
On Tue, Feb 2, 2010 at 3:01 AM, Madhavi Bhave wrote:
Dear R helpers,
Yesterday I had raised following query which was addressed by Mr Ellison. The
query and the wonderful solution as provided by Mr. Ellison are as given below.
## PROBLEM
I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
Its a $1000 face value, 3 year bond with 10% annual coupon and is priced at
101. The yield to maturity can be calculated after solving the equation -
1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
This can be solved by trial and error method s.t. ytm = 9.601%. I wanted to
find out how to solve this equation in R.
## SOLUTION
Mr. Elisson had given me following wonderful solution
f.ytm<-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
And I got the answer as 9.601.
## _
I was just trying to generalize this solution to any equation and accordingly
written a code as given below.
The following input I will be reading using csv file and thus my equation will
change if tenure or no_comp etc. changes. So taking into account the variable
nature of the input, I am trying to write a generalized code.
## Input
price = 101 # Price of bond
tenure = 3
no_comp = 1 # no of times coupon paid in a year.
coupon_rate = 0.10 # i.e. 10%
face_value = 100
# Computations
coupon_payment = face_value * coupon_rate
cash_flow = c(rep(c(coupon_payemnt), (no_comp * tenure - 1)), face_value +
coupon_payment)
cash_flow
## I am trying to customize the code as given by Mr Ellison.
f.ytm = function(ytm)
{
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value + coupon_payment)/((1+ytm)^(tenure * no_comp))) -
price
}
}
For this to work, tenure, no_comp, cash_flow, face_value and coupon_payment
have to be
visible to the function - i.e., they either have to be in the function's
calling environment
or in the global environment. These are called 'free variables' under the
lexical
scoping rules of R. (Welcome to function writing :)
You might want to look a little more closely at uniroot(), especially the
conventions
it requires for the *functions* it can evaluate. You want the body of what you
send
to uniroot() for evaluation to be a function of a single variable. Your
previously supplied
solution meets that requirement. This one doesn't (yet).
Moreover, it appears that E[i] is never used and nothing is returned from
f.ytm. I'd
suggest an excursion into R function writing fundamentals. Start with Ch. 10 of
the Introduction to R manual.
HTH,
Dennis
solution = uniroot(f.ytm, interval=c(0,25))
ytm = solution$root
However, when I execute this code I get following error.
> solution = uniroot(f.ytm, interval=c(0,25))
Error in uniroot(f.ytm, interval = c(0, 25)) : f.lower = f(lower) is NA
Please guide. ytm should be 0.09601 (i.e. 9.601%)
with regards
Madhavi Bhave
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[R] Yield to Maturity using R
Dear R helpers,
Yesterday I had raised following query which was addressed by Mr Ellison. The
query and the wonderful solution as provided by Mr. Ellison are as given below.
## PROBLEM
I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
Its a $1000 face value, 3 year bond with 10% annual coupon and is priced at
101. The yield to maturity can be calculated after solving the equation -
1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
This can be solved by trial and error method s.t. ytm = 9.601%. I wanted to
find out how to solve this equation in R.
## SOLUTION
Mr. Elisson had given me following wonderful solution
f.ytm<-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
And I got the answer as 9.601.
## _
I was just trying to generalize this solution to any equation and accordingly
written a code as given below.
The following input I will be reading using csv file and thus my equation will
change if tenure or no_comp etc. changes. So taking into account the variable
nature of the input, I am trying to write a generalized code.
## Input
price = 101 # Price of bond
tenure = 3
no_comp = 1 # no of times coupon paid in a year.
coupon_rate = 0.10 # i.e. 10%
face_value = 100
# Computations
coupon_payment = face_value * coupon_rate
cash_flow = c(rep(c(coupon_payemnt), (no_comp * tenure - 1)), face_value +
coupon_payment)
cash_flow
## I am trying to customize the code as given by Mr Ellison.
f.ytm = function(ytm)
{
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value + coupon_payment)/((1+ytm)^(tenure * no_comp))) -
price
}
}
solution = uniroot(f.ytm, interval=c(0,25))
ytm = solution$root
However, when I execute this code I get following error.
> solution = uniroot(f.ytm, interval=c(0,25))
Error in uniroot(f.ytm, interval = c(0, 25)) : f.lower = f(lower) is NA
Please guide. ytm should be 0.09601 (i.e. 9.601%)
with regards
Madhavi Bhave
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Re: [R] 'R' and 'Yield to Maturity'
Dear Sir,
That was GREAT!!!. Thanks a lot for the solution. Once again it showed how
powerful 'R' is otherwise I was breaking my head on Newton-Raphson method.
Thanks again Sir. That was really superb.
Madhavi
--- On Mon, 1/2/10, S Ellison wrote:
From: S Ellison
Subject: Re: [R] 'R' and 'Yield to Maturity'
To: "Craig P. Pyrame" , "Madhavi Bhave"
Cc: r-help@r-project.org
Date: Monday, 1 February, 2010, 3:41 AM
If you know the likely range, uniroot would do it.
f.ytm<-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
>>> "Craig P. Pyrame" 01/02/2010 10:19 >>>
Madhavi Bhave wrote:
> Dear R helpers
>
> I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
>
> Its a $1000 face value, 3 year bond with 10% annual coupon and is
priced at 101. The yield to maturity can be calculated after solving the
equation -
>
> 1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
>
> This can be solved by trial and error method s.t. ytm = 9.601%.
>
Why don't you use sage, for example:
sage: var('ytm');
sage: eqn = 1010 == (100/(1+ytm) + 100/(1+ytm)^2 + 1100/(1+ytm)^3);
sage: [solution.right().n() for solution in solve(eqn, ytm)][2]
0.0960070882662794
The third value is the only real solution.
There may be an R package for doing that, but I don't know one.
Best regards,
Craig
>
> My query is (1) if there is any R package which will calcualte ytm or
(2) is there any method in 'R' which can solve the above equation.
>
> Thanking you all in advance
>
> Regards
>
> Madhavi
>
>
>
>
>
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[R] 'R' and 'Yield to Maturity'
Dear R helpers I am calculating the 'Yield to Maturity' for the Bond with following characteristics. Its a $1000 face value, 3 year bond with 10% annual coupon and is priced at 101. The yield to maturity can be calculated after solving the equation - 1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3] This can be solved by trial and error method s.t. ytm = 9.601%. My query is (1) if there is any R package which will calcualte ytm or (2) is there any method in 'R' which can solve the above equation. Thanking you all in advance Regards Madhavi Your Mail works best with the New Yahoo Optimized IE8. Get it NOW! http://downloads.yahoo.com/in/internetexplorer/ [[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 sort data.frame
Dear R heleprs Suppose I have following data Scenarios combination_names series1 series2 Sc1 MAT2 GAU1 7.26554 8.409778 Sc2 MAT2 GAU2 7.438128 8.130275 Sc3 MAT3 GAU1 8.058422 8.06457 Sc4 MAT1 GAU2 8.179855 8.022071 Sc5 MAT3 GAU2 8.184033 8.191831 Sc6 MAT3 GAU2 7.50312 8.232425 Sc7 MAT1 GAU2 7.603291 8.200993 Sc8 MAT1 GAU1 8.221755 8.380097 Sc9 MAT3 GAU2 7.904908 8.088824 Sc10 MAT1 GAU3 7.67034 8.46376 I wish to sort thise data frame based on combination_names. Actually this is just an indicative data. I am deling with the data haveing 5000+ records. I just need to find out how to sort this data s.t, I will get say following result Scenarios combination_names series1 series2 Sc8 MAT1 GAU1 8.221755 8.380097 Sc4 MAT1 GAU2 8.179855 8.022071 Sc7 MAT1 GAU2 7.603291 8.200993 Sc10 MAT1 GAU3 7.67034 8.46376 and so I tried to understand the examples given in ?base::order but couldn't cracke it. Please guide Madhavi The INTERNET now has a personality. YOURS! See your Yahoo! Homepage. [[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] Please Please Please Help me!!
Dear R helpers
(I have already written the required R code which is giving me correct results
for a given single set of data. I just wish to wish to use it for multiple
data.)
I have defined a function (as given below) which helps me calculate Macaulay
Duration and Modified Duration and its working fine with given set of data.
My Code -
## ONS - PPA
duration = function(par_value, coupon_rate, frequency_copoun, tenure, ytm)
{
macaulay_duration = NULL
modified_duration = NULL
freq_coupon_new = NULL
if(frequency_copoun <= 0)
{
freq_coupon_new = 365
}
if(frequency_copoun > 0 & frequency_copoun <= 1)
{
freq_coupon_new = 12
}
if(frequency_copoun > 1 & frequency_copoun <= 2)
{
freq_coupon_new = 4
}
if(frequency_copoun > 2 & frequency_copoun <= 3)
{
freq_coupon_new = 2
}
if(frequency_copoun > 3 & frequency_copoun <= 4)
{
freq_coupon_new = 1
}
## COMPUTATIONS
terms_coupon_payment = (seq(1/freq_coupon_new, tenure, by =
1/freq_coupon_new))*freq_coupon_new
coupon = coupon_rate*par_value/100
coupon_amount = coupon/(freq_coupon_new)
cash_flow1 = rep(c(coupon_amount), (tenure*freq_coupon_new -
1))
cash_flow2 = par_value + coupon_amount
cash_flow = c(cash_flow1, cash_flow2)
ytm_effective = ((1+ytm/100)^(1/freq_coupon_new))-1
pv = NULL
for (i in 1:(tenure*freq_coupon_new))
{
pv[i] = cash_flow[i] / ((1+ytm_effective)^terms_coupon_payment[i])
}
macaulay_duration = sum(pv*terms_coupon_payment)/sum(pv)
modified_duration = macaulay_duration / (1+(ytm_effective)/freq_coupon_new)
return(data.frame(macaulay_duration, modified_duration))
}
### For a given data say
result = duration(par_value = 1000, coupon_rate = 10, frequency_copoun = 0,
tenure = 5, ytm = 12)
I get the output as
macaulay_duration modified_duration
1 1423.797 1423.795
## __
## MY PROBLEM
If instead of having only one set of data, suppose I have multiple data (say as
given below in a csv file), I am not able to get these results.
Suppose my 'input.csv' file is as given below.
par_value coupon_rate frequency_copoun tenure ytm
1000 10 0 5
12
100
7 1 8 11
Sir, I am not asking for the modification of existing code as it is running
fine with a single set of data (and I have checked that the output tallies with
other methods). I just want to use this code for multiple data so that I get an
output something like
maculay_duration modified_duration
1423.797 1423.795
44.339 44.307
I request you to please guide me. I undesratnd its not my right to seek help,
but this is for the third time I am requesting to guide me.
Thanking you all
Regards
Madhavi
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[R] Macualay Duration code in a Functional Form - Please Help
# I have written this code in Notepad++ and copied here.
## ONS - PPA
Duration = function(par_value, coupon_rate, freq_coupon, tenure, ytm)
{
macaulay_duration = NULL
modified_duration = NULL
freq_coupon_new = NULL
if(freq_coupon <= 0)
{
freq_coupon_new = 365
}
if(freq_coupon > 0 & freq_coupon <= 1)
{
freq_coupon_new = 12
}
if(freq_coupon > 1 & freq_coupon <= 2)
{
freq_coupon_new = 4
}
if(freq_coupon > 2 & freq_coupon <= 3)
{
freq_coupon_new = 2
}
if(freq_coupon > 3 & freq_coupon <= 4)
{
freq_coupon_new = 1
}
## COMPUTATIONS
terms_coupon_payment = (seq(1/freq_coupon_new, tenure, by =
1/freq_coupon_new))*freq_coupon_new
coupon = coupon_rate*par_value/100
coupon_amount = coupon/(freq_coupon_new)
cash_flow1 = rep(c(coupon_amount), (tenure*freq_coupon_new -
1))
cash_flow2 = par_value + coupon_amount
cash_flow = c(cash_flow1, cash_flow2)
ytm_effective = ((1+ytm/100)^(1/freq_coupon_new))-1
pv = NULL
for (i in 1:(tenure*freq_coupon_new))
{
pv[i] = cash_flow[i] / ((1+ytm_effective)^terms_coupon_payment[i])
}
macaulay_duration = sum(pv*terms_coupon_payment)/sum(pv)
modified_duration = macaulay_duration / (1+(ytm_effective)/freq_coupon_new)
return(data.frame(macaulay_duration, modified_duration))
}
result = Duration(par_value = 1000, coupon_rate = 10, freq_coupon = 0, tenure =
5, ytm = 12)
## ___
When I run this function, I get the values of Macaulay Duration and Modified
Duration
> result
macaulay_duration modified_duration
1 1423.797 1423.795
### MY PROBLEM
I have arrived at a result using only one set of observations i.e. for the
following data -
Duration(par_value = 1000, coupon_rate = 10, freq_coupon = 0, tenure = 5, ytm =
12)
However, if I need to obtain these results for multiple records, how do I
calculate and obtain the result in a tabular form?
e.g. suppose my input data file is 'instrument details.csv' given as
id par_value coupon_rate frequency_coupon tenure ytm
1 1000 10 0
5 12
2 100 7 1
8 11
### frequency_coupon is coded s.t. if frequency_coupon = 0, no of compoundings
in a year = 365 and if it is 1, then no of compoundings = 12
Then how do modify the above code?
I have tried to convert in a matrix form as follows
I have added following code after the function is defined i.e. after
#return(data.frame(macaulay_duration, modified_duration))
#}
# Added code
ONS = read.csv('instrument details.csv')
n = length(ONS$par_value)
par_value = matrix(data = ONS$par_value, nrow = n, ncol = 1, byrow = TRUE)
coupon_rate = matrix(data = ONS$coupon_rate, nrow = n, ncol = 1, byrow =
TRUE)
freq_coupon = matrix(data = ONS$frequency_copoun, nrow = n, ncol = 1, byrow =
TRUE)
tenure = matrix(data = ONS$tenure, nrow = n, ncol = 1, byrow = TRUE)
ytm = matrix(data = ONS$ytm, nrow = n, ncol = 1, byrow = TRUE)
result = matrix(data = NA, nrow = n, ncol = 2, byrow = TRUE)
result = Duration(par_value, coupon_rate, freq_coupon, tenure, ytm)
##
When I run result, besides getting 50 warnings, I get following
> result
macaulay_duration modified_duration
1 826.9026 826.9019
2 826.9026 826.9019
which is I know wrong.
Is there any other way I can use the function defined above to process multiple
recrds.
Thanking you and sincerely apologize for writing such a long mail as I wanted
to be clear in my communication.
Regards
Madhavi Bhave
ONS = read.csv('instrument details.csv')
n = length(ONS$par_value)
par_value = matrix(data = ONS$par_value, nrow = n, ncol = 1, byrow = TRUE)
coupon_rate = matrix(data = ONS$coupon_rate, nrow = n, ncol = 1, byrow =
TRUE)
freq_coupon = matrix(data = ONS$frequency_copoun, nrow = n, ncol = 1, byrow =
TRUE)
tenure = matrix(data = ONS$tenure, nrow = n, ncol = 1, byrow = TRUE)
ytm = matrix(data = ONS$ytm, nrow = n, ncol = 1, byrow = TRUE)
result = matrix(data = ONS$par_value, nrow = n, ncol = 2, byrow =
TRUE)
result = Duration(par_value, coupon_rate, freq_coupon, tenure, ytm)
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[R] Macaulay Duration for Group
Dear R helpers
I have following csv file which is an input
id par_value coupon_rate frequency_coupon tenure ytm
1 1000 10 1
5 12
# Here frequency_coupon is coded s.t. 0 means Daily compounding, 1 means
monthly compouding, 2 means Quarterly, 3 means Half yearly and 4 means
only once. Thus in the case the frequency_coupon = 1 means, total number of
times compounding is done = 12.
My R Code for calcualting Macaulay Duration is as follows -
## INPUT
ONS = read.csv('instrument details..csv')
par_value = ONS$par_value
coupon = ONS$coupon_rate*par_value/100
freq_coupon = ONS$frequency_copoun
tenure = ONS$tenure
ytm = ONS$ytm
#
_
## COMPUTATIONS
macaulay_duration = NULL
modified_duration = NULL
freq_coupon_new = NULL
if(freq_coupon <= 0)
{
freq_coupon_new = 365
}
if(freq_coupon > 0 & freq_coupon <= 1)
{
freq_coupon_new = 12
}
if(freq_coupon > 1 & freq_coupon <= 2)
{
freq_coupon_new = 4
}
if(freq_coupon > 2 & freq_coupon <= 3)
{
freq_coupon_new = 2
}
if(freq_coupon > 3 & freq_coupon <= 4)
{
freq_coupon_new = 1
}
## COMPUTATIONS
terms_coupon_payment = (seq(1/freq_coupon_new, tenure, by =
1/freq_coupon_new))*freq_coupon_new
coupon_amount = coupon/(freq_coupon_new)
cash_flow1 = rep(c(coupon_amount), (tenure*freq_coupon_new -
1))
cash_flow2 = par_value + coupon_amount
cash_flow = c(cash_flow1, cash_flow2)
ytm_effective = ((1+ytm/100)^(1/freq_coupon_new))-1
pv = NULL
for (i in 1:(tenure*freq_coupon_new))
{
pv[i] = cash_flow[i] / ((1+ytm_effective)^terms_coupon_payment[i])
}
macaulay_duration = sum(pv*terms_coupon_payment)/sum(pv)
modified_duration = macaulay_duration / (1+(ytm_effective)/freq_coupon_new)
macaulay_duration
modified_duration
## _
# My PROBLEM
Here I am dealing with only one id i.e. only one record. However, if Instead of
one record, ahve say 20 records, how do I calculate the Macaulay Duration for
each of these 20 records. One option is to run this code 20 times *which I
guess will be foolish thing to do. Other method is to define above code as some
function and tehn run this function for each of these records, but I don't
underatnd how to write a function and thord option is to treat the input of
these 20 records in a matrix form, which I had tried unsuccessfully.
Please guide me as to how do I modify the R-code to calculate Mac duration for
each of tehse records and store tehm.
Regards
Madhavi Bhave
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[R] Parameters of Logistic Distribution and (3 Parameter) Log Logistic Distribution
Dear R Helpers Please guide me how one can estimate the parameters of Logistic Distribution and 3 Parameter Log-logistic distribution for a given data. data <- c(2987.43,2990.12,3023.52,2964.79,3019.60,3051.07,3080.16,2944.15,3035.19,3023.46,2985.05,2970.95,3192.36,3084.39,2926.23,2952.15,3064.15,3003.20,2980..35,2980.45,3043.12,3115.53,3006.90,2946.03,3039.97,3064.01,3000.56,3049.57,3042.54,3037.63,2982.03,2889.74,3043.83,2930.95,3020.65,3009.21,3084.16,2954.05,2991.04,3083.10,3007.26,2949.58,2995.65,3078.36,3031.64,3001.28,3103.32,3015.04,2994.45,2963.71,2932.90,3021.31,3074.72,2980.15,3002.29,3088.18,2991.39,2942.90,3057.91,3023.25,3192.67,2966.49,3049.31,2915.38,3045.27,2852.72,2999.25,2978.52,3040.07,2945.50,3047.47,2915.95,3012.24,2985.80,2971.04,3035.72,3025.40,3014.76,2979.62,3029.20,2938.38,2966.47,3017.81,3016.43,2989.60,2941.22,3038.30,3033.44,3003.77,2950.02,3053.19,3011.69,2916.34,2918..10,3049.98,3062.46,2948.55,3072.90,3113.52,2987.61) Thanking in advance. With regards Madhavi Love Cricket? Check out live scores, photos, video highlights and more. Click here http://cricket.yahoo.com [[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] Problem with Poisson - Chi Square Goodness of Fit Test - New Mail
Dear R-help, Chi Square Test for Goodness of Fit I have got a discrete data as given below (R script) No_of_Frauds<-c(1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,1,1,1,1,3,2,1,1,1,2,1,1,2,1,1,1,1,1,2,1,3,1,2,1,2,14,2,1,1,38,3,3,2,44,1,4,1,4,1,2,2,1,3) I am trying to fit Poisson distribution to this data using R. My R script is as under : # R SCRIPT for Fitting Poisson Distribution No_of_Frauds<-c(1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,2,2,2,1,1,2,1,1,1,1,4,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,1,1,1,1,3,2,1,1,1,2,1,1,2,1,1,1,1,1,2,1,3,1,2,1,2,14,2,1,1,38,3,3,2,44,1,4,1,4,1,2,2,1,3) N <- length(No_of_Frauds) Average <- mean(No_of_Frauds) Lambda <- Average i <- c(0:(N-1)) pmf <- dpois(i, Lambda, log = FALSE) # # Ho: The data follow Poisson Distribution Vs H1: Not Ho # observed frequencies (Oi) variable.cnts <- table(No_of_Frauds) variable.cnts.prs <- dpois(as.numeric(names(variable.cnts)), lambda) variable.cnts <- c(variable.cnts, 0) variable.cnts.prs <- c(variable.cnts.prs, 1-sum(variable.cnts.prs)) tst <- chisq.test(variable.cnts, p=variable.cnts.prs) chi_squared <- as.numeric(unclass(tst)$statistic) p_value <- as.numeric(unclass(tst)$p.value) df <- tst[2]$parameter cv1 <- qchisq(p=.01, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE) cv2 <- qchisq(p=.05, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE) cv3 <- qchisq(p=.1, df=tst[2]$parameter, lower.tail = FALSE, log.p = FALSE) #- # Expected value # variable.cnts.prs * sum(variable.cnts) # if tst > cv reject Ho at alpha confidence level #- if(chi_squared > cv1) Conclusion1 <- 'Sample does not come from the postulated probability distribution at 1% los' else Conclusion1 <- 'Sample comes from postulated prob. distribution at 1% los' if(chi_squared > cv2) Conclusion2 <- 'Sample does not come from the postulated probability distribution at 5% los' else Conclusion2 <- 'Sample comes from postulated prob. distribution at 1% los' if(chi_squared > cv3) Conclusion3 <- 'Sample does not come from the postulated probability distribution at 10% los' else Conclusion3 <- 'Sample come from postulated prob distribution at 1% los' #- # Printing RESULTS print(chi_squared) print(p_value) print(df) print(cv1) print(cv2) print(cv3) print(Conclusion1) print(Conclusion2) print(Conclusion3) # End of R Script Problem Faced : When I run this script using R – console, I am getting value of Chi – Square Statistics as high as “6.95753e+37” When I did the same calculations in Excel, I got the Chi Square Statistics value = 138.34. Although it is clear that the sample data doesn’t follow Poisson distribution, and I will have to look for other discrete distribution, my problem is the HIGH Value of Chi Square test statistics. When I analyzed further, I understood the problem. (A) By convention, if your Expected frequency is less than 5, then by we put together such classes and form a new class such that Expected frequency is greater than 5 and also accordingly adjust the observed frequencies. X Oi Ei ((Oi - Ei)^2)/Ei 0 0 10 9.96 1 72 23 103.79 2 17 27 3.54 3 5 21 11.85 4 3 12 6.71 5 4 9 2.51 Total 101 101 138.34 When I apply this logic in Excel, I am getting the reasonable result (i.e. 138.34), however in Excel also, if I don’t apply this logic, my Chi square test statistic value is as high as 4.70043E+37. My question is how do I modify my R – script, so that the logic mentioned in (A) i.e. adjusting the Expected frequencies (and accordingly Observed frequencies) is applied so that the expected frequency becomes greater than 5 for a given class, thereby resulting in reasonable value of Chi Square test Statistics. I am also attaching the xls file for
[R] Interpreting Logistic Regression
Hi ! This is Madhavi from Mumbai, India. Incidently this is my first post. I am working on Credit Scoring Model and using R, I have run the logistic regression. I have received following Output. I have two questions (a) What is the significance of "family = binomial(link = logit)". Why do I have to mention Binomial? Is it because my dependent variable assumes only two values 0 and 1? Can I write name of some other Statistical distribution (say Poisson or Negative Binomial) in place of Binomial? How will it affect my results? (b) How do I interpret the "R" result as given below? I know all the variables are significant. How do I get Log Likelihood ratio, Odds ratio etc.? Please can anyone help me out. With warm regards Madhavi R OUTPUT Call: glm(formula = Y ~ Age1 + Age2 + Sex + Education + Profession + SavingsAccount +    CurrentAccount, family = binomial(link = logit), data = ons) Deviance Residuals:     Min       1Q   Median       3Q      Max  -3.21142 -0.42556 -0.15911 -0.02954  3.02465  Coefficients:                  Estimate Std. Error z value Pr(>|z|)   (Intercept)      2.627725  0.110752 23.726 < 2e-16 *** Age1             0.692180  0.070410  9.831 < 2e-16 *** Age2            -2.817883  0.080801 -34.874 < 2e-16 *** Sex             -0.486132  0.049766 -9.768 < 2e-16 *** Education       -0.682142  0..046507 -14.667 < 2e-16 *** Profession      -0.690937  0.069032 -10.009 < 2e-16 *** SavingsAccount  -1.891455  0.074906 -25.251 < 2e-16 *** CurrentAccount  -1.367460  0.079604 -17.178 < 2e-16 *** --- Signif. codes: 0 â***â 0.001 â**â 0.01 â*â 0.05 â.â 0.1 â â 1 (Dispersion parameter for binomial family taken to be 1)    Null deviance: 26932 on 24999 degrees of freedom Residual deviance: 14615 on 24983 degrees of freedom  (2 observations deleted due to missingness) AIC: 14649 Number of Fisher Scoring iterations: 6 Unlimited freedom, unlimited storage. Get it now, on http://help.yahoo.com/l/in/yahoo/mail/yahoomail/tools/tools-08.html/ [[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.
