Dear R helpers,
(At the outset I sincerely apologize if I have not put forward my following
query properly, though I have tried to do so.)
Following is a curtailed part of my R - code where I am trying to generate say
100 random no.s for each of the products under consideration.
library(plyr
Dear Mr Weylandt and R helpers,
Thanks a lot for your suggestion. Unfortunately the "return" statement in my
original R code returns me different results which are obtained after
processing the function I have constructed.
My requirement for storing the product-wise random numbers is just a pa
Dear R helpers
I have following two data.frames viz. equity_data and param.
equity_data = data.frame(security_id = c("Air", "Air", "Air", "Air", "Air",
"Air", "Air", "Air", "Air", "Air", "Air", "Air", "AB", "AB", "AB", "AB", "AB",
"AB", "AB", "AB", "AB", "AB", "AB", "AB", "AD", "AD", "AD", "AD
Dear R helpers
I have a data which gives Month-wise and Rating-wise Rates. So the input file
is something like
month rating rate
January AAA 9.04
February AAA 9.07
..
.
: Re: [R] One silly question about "tapply output"
To: "Vincy Pyne"
Received: Wednesday, October 27, 2010, 11:28
AM
Assign your result to an object and then write out the object as a csv
file. For example:
x<-data.frame(rating=rep(letters[1:3],2),rate=runif(1:6)) # example
Dear R helpers
I have following input
f = c(257, 520, 110). I need to generate a decreasing sequence (decreasing by
100) which will give me an input (in a tabular form) like
257, 157, 57
520, 420, 320, 220, 120, 20
110, 10
I tried the following R code
f = c(257, 520, 110)
yy = matrix(da
rbing you Sir and hope I am able to put up my
problem in a proper manner.
Regards
Vincy Pyne
--- On Thu, 12/9/10, Jan van der Laan wrote:
From: Jan van der Laan
Subject: Re: [R] Sequence generation in a table
To: r-help@r-project.org, vincy_p...@yahoo.ca
Received: Thursday, December 9, 2010,
, Petr PIKAL wrote:
From: Petr PIKAL
Subject: Re: [R] Sequence generation in a table
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Thursday, December 9, 2010, 12:03 PM
Hi
r-help-boun...@r-project.org napsal dne 09.12.2010 12:41:47:
> Dear Sir,
>
> Sorry to bother y
n I tried
mydat.sort <- mydat[order(mydat$date)]
> mydat.sort <- mydat[order(mydat$date)]
Error in `[.data.frame`(mydat, order(mydat$date)) : undefined columns selected
Kindly guide
Vincy Pyne
[[alternative HTML version deleted]]
___
Dear sir,
Thanks a lot for your great guidance. It worked fantastically.
Regards
Vincy Pyne
--- On Thu, 12/30/10, Henrique Dallazuanna wrote:
From: Henrique Dallazuanna
Subject: Re: [R] Sorting data.frame datewise in a descending order
To: "Vincy Pyne"
Cc: r-help@r-project.or
Dear R helpers
Wish you all a very Happy and Prosperous New Year 2011.
I have following query.
country = c("US", "France", "UK", "NewZealand", "Germany", "Austria", "Italy",
"Canada")
Through some other R process, the result.csv file is generated as
result.csv
var1 var2 var3 var4
Dear R helpers
I have a dataframe as
df = data.frame(x = c(1, 14, 3, 21, 11), y = c(102, 500, 40, 101, 189))
> df
x y
1 1 102
2 14 500
3 3 40
4 21 101
5 11 189
# Actually I am having dataframe having multiple columns. I am just giving an
example.
I need to subtract all the rows of df
Hi
I have a large credit portfolio (exceeding 5 borrowers). For particular
process I need to add up the exposures based on the bands. I am giving a small
test data below.
rating <- c("A", "AAA", "A", "BBB","AA","A","BB", "BBB", "AA", "AA", "AA", "A",
"A", "AA","BB","BBB","AA", "A", "AAA","
learn something from THE WISE like you.
Thanks once again Sir. Your help was great and it means a lot to me and for
other freshers like me.
Regards
Vincy Pyne
--- On Fri, 8/27/10, David Winsemius wrote:
From: David Winsemius
Subject: Re: [R] Band-wise Sum
To: "Vincy Pyne"
Cc
sorry for any mis-communication in my earlier mail. I could test
the reply sent to me earlier by Winsemius Sir only today as I was traveling
over weekends. Also, I have tried to go through earlier emails dealing with
such conditional sums. Unfortunately, I couldn't understand as I have rec
HTH,
Dennis
--- On Mon, 8/30/10, David Winsemius wrote:
From: David Winsemius
Subject: Re: [R] Band-wise Conditional Sum - Actual problem
Cc: r-help@r-project.org
Received: Monday, August 30, 2010, 2:43 PM
On Aug 30, 2010, at 4:05 AM, Vincy Pyne wrote:
> Dear R helpers,
>
> Th
Hi!
I wish to know if there is any R forum which is meant only for Statistics? I
mean where we can clarify our statistics doubts and seek knowledge. I know
there are lot many books and internet sites, but 'R forum' has altogether
different standard and very high level and one can learn a lot fr
Dear R helpers,
I have one basic doubt about the value of pi. In school, we have learned that
pi = 22/7 (which is = 3.142857). However, if I type pi in R, I get pi =
3.141593. So which value of pi should be considered?
Regards
Vincy
[[alternative HTML version deleted]]
_
That's the beauty of this R forum. This forum is full of knowledgeable wizards
and replies received along-with the related discussions pertaining to a simple
harmless question like this enriches us tremendously. Thanks a lot for all your
comments. I am sticking to the value of 'pi' as provided i
Dear R helpers
I seem to have one trivial problem but can't find solution to it.
Suppose I have following input.
A = c(1, 3, 0, 5, 8) # 3rd element is 0
B = c(100, 30, 0, 25, 40) # 3rd element is 0
C = A/B
> C
[1] 0.01 0.10 NaN 0.20 0.20
Obviously, I can't di
Dear R helpers,
Suppose I have following data.frame.
df <- data.frame(category = c("treat_A", "treat_A", "treat_A", "treat_A",
"treat_A", "treat_A", "treat_A", "treat_A", "treat_B", "treat_B", "treat_B",
"treat_B", "treat_B", "treat_B", "treat_B", "treat_B"), type = c("AA", "",
"B", "AAA",
Dear R helpers,
xx = data.frame(country = c("USA", "UK", "Canada"), x = c(10, 50, 20), y =
c(40, 80, 35), z = c(70, 62, 10))
> xx
country x y z
1 USA 10 40 70
2
UK 50 80 62
3 Canada 20 35 10
I need to arrange this as a new data
Dear R helpers
Suppose
val1 = c(10, 20, 35, 80, 12)
val2 = c(3, 8, 11, 7)
I want to select either val1 or val2 depending on value of third quantity val3.
val3 assumes either of the values "Monthly" or "Yearly".
If val3 = "Monthly", then val = val1 and if val3 = "Yearly", then val = val2.
I tr
t; if (val3 == "Monthly") val <- val1
> else val <- val2
>
> For a simple calculation like this I'd probably use the former; if the
> calculation got more complex I'd prefer the latter.
>
> Duncan Murdoch
>
>>
>> Single "=" is f
Dear R helpers
I have following data.frame and for each product_name, I have associated mean
and standard deviation. I need to generate 1000 random no.s for each of these
products and find the respective mean and standard deviation.
My R code is as follows.
library(plyr)
library(reshape2)
Dear R helpers
Suppose I have a data frame as given below
mydat = data.frame(x = c(1,1,1, 2, 2, 2, 2, 2, 5, 5, 6), y = c(10, 10, 10, 8,
8, 8, 7, 7, 2, 2, 4))
mydat
x y
1 1 10
2 1 10
3 1 10
4 2 8
5 2 8
6 2 8
7 2
Thanks sir for your reply. Unfortunately I couldn't figure out the solution.
Vincy
--- On Sat, 3/12/11, Dennis Murphy wrote:
From: Dennis Murphy
Subject: Re: [R] Identifying unique pairs
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Saturday, March 12, 2011, 11:45 AM
Hi
3/12/11, Petr Savicky wrote:
From: Petr Savicky
Subject: Re: [R] Identifying unique pairs
To: r-help@r-project.org
Received: Saturday, March 12, 2011, 2:10 PM
On Sat, Mar 12, 2011 at 03:20:01AM -0800, Vincy Pyne wrote:
> Dear R helpers
>
> Suppose I have a data frame as given below
>
&
Dear R helpers
Suppose I have a vector as
vect_1 = c("AAA", "AA", "A", "BBB", "BB", "B", "CCC")
vect_1_id = c(1:length(vect_1))
Through some process I obtain
vect_2_id = c(2, 3, 7), then I need a new vector say vect_2 which will give me
vect2 = ("AA", "A", "CCC") i.e. I need the subset of ve
Dear R helpers
Suppose,
x = c(0, 1, 2, 3)
y = c("A", "B", "C", "D")
z = c(1, 3)
For given values of z, I need to the values of y. So I should get "B" and "D".
I tried doing
y[x][z] but it gives
> y[x][z]
[1] "A" "C"
Kindly guide.
Regards
Vincy
[[alternative HTML version
Dear R helpers,
Suppose I have stock returns data of say 1500 companies each for say last 4
years. Thus I have a matrix of dimension say 1000 * 1500 i.e. 1500 columns
representing companies and 1000 rows of their returns.
I need to find the correlation matrix of these 1500 companies.
So I can
refer to the packages you have suggested.
Thanks again
Vincy
--- On Mon, 3/21/11, Peter Langfelder wrote:
From: Peter Langfelder
Subject: Re: [R] Correlation for no of variables
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Monday, March 21, 2011, 4:50 PM
On Mon, Mar 21, 2011
Dear R helpers
exposure <- data.frame(id =
c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20),
ead = c(9483.686,5,6843.4968,10509.37125,21297.8905,5,706152.8354,
62670.5625, 687.801995,50641.4875,59227.125,43818.5778,52887.72534,601788.7937,
56813.14859,4012356.056,1419501.179,210853
will be executed outside the loop.
Once this is generated, writing of the csv file should not be problem outside
the loop.
Regards
Vincy
--- On Fri, 3/25/11, Ista Zahn wrote:
From: Ista Zahn
Subject: Re: [R] Appending data to a data.frame and writing a csv
To: "Vincy Pyne"
Cc
Dear R helpers
Suppose I have a data.frame as given below -
my_dat = data.frame(class = c("XYZ", "XYZ", "XYZ", "XYZ", "XYZ","ABC", "ABC",
"ABC", "ABC", "ABC" ), var1 = c(20, 14, 89, 81, 17, 44, 36, 41, 11, 36), var2
= c(1001, 250, 456, 740, 380, 641, 111, 209, 830, 920))
> my_dat
class var
Dear R helpers
I am resending my mail as the output I desire was not properly visible and I
apologize for the same.
Suppose I have a data.frame as given below -
my_dat = data.frame(class = c("XYZ", "XYZ", "XYZ", "XYZ", "XYZ","ABC", "ABC",
"ABC", "ABC", "ABC" ), var1 = c(20, 14, 89, 81, 17, 44
Dear sir,
Thanks for the great solution.
Regards
Vincy
--- On Mon, 3/28/11, Henrique Dallazuanna wrote:
From: Henrique Dallazuanna
Subject: Re: [R] Ordering data.frame based on class
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Monday, March 28, 2011, 9:02 PM
Try thi
Dear R helpers
Suppose I have a vector as
vect1 = as.character(c("ABC", "XYZ", "LMN", "DEF"))
> vect1
[1] "ABC" "XYZ" "LMN" "DEF"
I want to reverse the order of this vector as
vect2 = c("DEF", "LMN", "XYZ", "ABC")
Kindly guide
Regards
Vincy
[[alternative HTML version deleted]]
Dear R list, I have one very elementary question regrading correlation between
two variables.
x = c(44,46,46,47,45,43,45,44)
y = c(44,43,41,41,46,48,44,43)
> cov(x, y)
[1] -2.428571
However, if I try to calculate the covariance using the formula as
covariance = sum((x-mean(x))*(y-mean(y)))/8
] Correlation discrepancy
To: r-help@r-project.org
Cc: "Vincy Pyne"
Received: Tuesday, August 23, 2011, 11:38 AM
In addition, something has gone wrong, Vincy, with your data x,y
between evaluating cov(x,y) and evaluating your explicit formula.
If I repeat your commands:
x = c(44,46,46,47,4
-project.org
Cc: "Vincy Pyne"
Received: Tuesday, August 23, 2011, 11:38 AM
In addition, something has gone wrong, Vincy, with your data x,y
between evaluating cov(x,y) and evaluating your explicit formula.
If I repeat your
commands:
x = c(44,46,46,47,45,43,45,44)
y = c(44,43,41,41,4
Dear R list
I am trying to understand the auto-correlation concept. Auto-correlation is the
self-correlation of random variable X with a certain time lag of say t.
The article
"http://www.mit.tut.fi/MIT-3010/luentokalvot/lk10-11/MDA_lecture16_11.pdf";
(Page no. 9 and 10) gives the methodology
: Prof Brian Ripley
Subject: Re: [R] Autocorrelation using library(tseries)
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received:
Wednesday, August 24, 2011, 9:08 AM
Your understanding is wrong. For a start, there is no function acf() in
package tseries: it is in stats.
And the au
Hi,
I have one basic doubt. Suppose X ~ N(50,10).
I need to calculate Probability X = 50.
dnorm(50, 50, 10) gives me
[1] 0.03989423
My understanding is (which is bit statistical or may be mathematical) on a
continuous scale, Probability of the type P(X = .) are nothing but
1/Infinity i.e.
Ellison and Mr Mark for your
guidance.
Regards
Vincy
--- On Wed, 9/14/11, S Ellison wrote:
From: S Ellison
Subject: RE: [R] Question regarding dnorm()
To: "Vincy Pyne" , "r-help@r-project.org"
Received: Wednesday, September 14, 2011, 11:37 AM
You have calculated dens
Dear R helpers,
I am using Beta distribution to generate the random no.s (recovery rates in my
example). However, each time I need to save these random no.s in a csv format.
To distinguish different csv files, one way I thought was use of Sys.time in
the file name. My code is as follows -
# My
Dear Mr Newmiller and Mr Oettli,
Thanks a lot for your valuable guidance. Task is done. Thanks again.
Regards
Vincy
--- On Wed, 7/4/12, Jeff Newmiller wrote:
From: Jeff Newmiller
Subject: Re: [R] How to use Sys.time() while writing a csv file name
To: "Vincy Pyne" , r-help@r-p
Dear R helpers,
I have a data.frame as given below -
dat1 = data.frame(date =
as.Date(c("3/30/12","3/29/12","3/28/12","3/27/12","3/26/12",
"3/23/12","3/22/12","3/21/12","3/20/12",
"3/30/12","3/29/12","3/28/12","3/27/12",
"3/26/12","3/23/12","3/22/12","3/21/12","3/20/12",
"3/30/12","3/29/12","3
command
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Wednesday, July 18, 2012, 11:18 AM
Hello,
Try the following.
# This is your code
df_sorted = df[order(as.Date(df$date, "%m/%d/%Y"), decreasing = TRUE),]
# This is my code
nams <- as.character(unique(dat1$nam
Dear R helpers,
I have one trivial problem while writing an output file in csv format.
I have two dataframes say df1 and df2 which I am reading from two different csv
files.
df1 has column names as date, r1, r2, r3 while the dataframe df2 has column
names as date, 1w, 2w.
(the dates in both
Dear R helpers,
I am not a statistician and right now struggling with Richards curve. Wikipedia
says
(http://en.wikipedia.org/wiki/Generalised_logistic_function)
The "generalized logistic curve or function", also known as Richard's curve is
a widely-used and flexible sigmoid function for growt
Dear r helpers
This may be very elementary question but I couldn't figure out what does the
operator %% do?
E.g.
p <- 100
q <- 200
p%%q
[1] 100
q%%p
[1] 0
Please guide.
Vincy
[[alternative HTML version deleted]]
__
R-help@r-project.org m
7/13/11, ONKELINX, Thierry wrote:
From: ONKELINX, Thierry
Subject: RE: [R] Meaning of "%%"
To: "Vincy Pyne" , "r-help@r-project.org"
Received: Wednesday, July 13, 2011, 10:13 AM
help("%%")
---
Dear R helpers,
I need to use KS and AD test for Generalized Pareto and Generalized extreme
value.
E.g. if I need to use KS for Weibull, I have teh syntax
ks.test(x.wei,"pweibull", shape=2,scale=1)
Similarly, for AD I use
ad.test(x, distr.fun, ...)
My problem is fir given data, I have estima
Dear R forum
I have two datafarmes with category and cat_val forming one dataframe and cust
and cust_category forming another dataframe.
category = c("C", "D", "B", "A")
cat_val = c(0.10, 0.25, 0.40, 0.54)
cust = c("cust_1", "cust_2", "cust_3", "cust_4", "cust_5", "cust_6", "cust_7",
"cust_8",
Dear R helpers,
Following is my R code where I am trying to calculate returns and then trying
to create a data.frame. Since, I am not aware how many instruments I will be
dealing so I have constructed a function. My R code is as follows -
library(plyr)
mydata <- data.frame(instru_name =
c("in
required data.frame.
Thanks again.
With warm regards
Vincy
--- On Wed, 1/11/12, jim holtman wrote:
From: jim holtman
Subject: Re: [R] Constructing a data.frame from csv files
To: "Vincy Pyne"
Cc: r-help@r-project.org
Received: Wednesday, January 11, 2012, 1:49 PM
The error message s
Dear R helpers
Suppose
x <- c(1:3)
y <- matrix(1:12, ncol = 3, nrow = 4)
> y
[,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
I wish to multiply 1st column of y by first element of x i.e. 1, 2nd column of
y by 2nd element of x i.e. 2 an so
quot;Vincy Pyne"
Cc: r-help@r-project.org
Received: Friday, April 20, 2012, 8:57 AM
try this:
x <- 1:3
y <- matrix(1:12, ncol = 3, nrow = 4)
y * rep(x, each = nrow(y))
I hope it helps.
Best,
Dimitris
On 4/20/2012 10:51 AM, Vincy Pyne wrote:
> Dear R helpers
>
> Suppose
&
Dear R helpers,
I have two separate data frames. In one data frame the transaction data is
stored and the other data frame has exchange rates stored say rate_A and rate_B
where rate_A and rate_B are series of rates.
rate_A and rate_B are properly defined and I am reading them through the
appr
Dear R helpers,
I have recently installed R version 2.15.0
I just wanted to calculate
mean(16, 18)
Surprisingly I got answer as
> mean(16, 18)
[1] 16
> mean(18, 16)
[1] 18
> mean(14, 11, 17, 9, 5, 18)
[1] 14
So instead of calculating simple Arithmetic average, mean command is generati
N?
To: "Vincy Pyne" , "r-help@r-project.org"
Received: Tuesday, May 22, 2012, 9:17 AM
You'll need to pass the data as a vector.
mean(16, 18) is asking the mean of 16. 18 is passed to the second argument
which is trim. So you are doing mean(16, trim = 18)
What you want is
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