[R] creating mulptiple new variables from one data.frame according to columns and rows in that frame
Hayes, Daniel
D.J.Hayes at liverpool.ac.uk
Wed Nov 4 17:44:16 CET 2009
YES, that does the trick.
Glad to have your help for I had no idea of the existence of FAQ 7.31 nor for that matter do I completely understand what floating point number are (but that is another story :P))
Think I am all set.
Cheers again for your time and energy.
Daniel
-----Original Message-----
From: jim holtman [mailto:jholtman at gmail.com]
Sent: 04 November 2009 15:30
To: Hayes, Daniel
Cc: r-help at lists.R-project.org
Subject: Re: [R] creating mulptiple new variables from one data.frame according to columns and rows in that frame
My guess is that we are being affected by FAQ 7.31 (good old floating
point numbers). The test 'age %in% 5:50' might be affected by round
off. Something like the following might be better:
age < 5 | (abs(age - round(age)) < 0.001)
This should give TRUE for all ages that are 'close' to the year. Take
a look at your data where you thing values might be missing and set
'options(digit=20)' to print out the full values.
On Wed, Nov 4, 2009 at 8:03 AM, Hayes, Daniel <D.J.Hayes at liverpool.ac.uk> wrote:
> Jim Holtman,
> Thank you for your reply.
> Your script is very concise and I think it could help me.
> However when I run it on my real data object (musigma.lat.m) the age range from 5-50 skips certain full years (see script below).
> Am not sure why that is and no error is given.
> Hoping you can help.
>
> Thank you in advance for your time and energy.
> All the best,
> Daniel
>
>> dput(musigma.lat.m[580:620,])
> structure(list(age = c(48.25, 48.3333333333333, 48.4166666666667,
> 48.5, 48.5833333333333, 48.6666666666667, 48.75, 48.8333333333333,
> 48.9166666666667, 49, 49.0833333333333, 49.1666666666667, 49.25,
> 49.3333333333333, 49.4166666666667, 49.5, 49.5833333333333, 49.6666666666667,
> 49.75, 49.8333333333333, 49.9166666666667, 50, 0, 0.0833333333333333,
> 0.166666666666667, 0.25, 0.333333333333333, 0.416666666666667,
> 0.5, 0.583333333333333, 0.666666666666667, 0.75, 0.833333333333333,
> 0.916666666666667, 1, 1.08333333333333, 1.16666666666667, 1.25,
> 1.33333333333333, 1.41666666666667, 1.5), country = structure(c(1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("Bolivia", "Brazil",
> "Colombia", "Dominican Rep.", "El Salvador", "Guatemala", "Guyana",
> "Haiti", "Honduras", "Nicaragua", "Paraguay", "Peru", "Suriname"
> ), class = "factor"), mu = c(10.7198320154036, 10.7193221119285,
> 10.7188036231439, 10.7182764259851, 10.7177406001273, 10.7171962535812,
> 10.7166435245754, 10.7160826629999, 10.7155141252060, 10.7149385933270,
> 10.7143568116012, 10.7137696820872, 10.7131779280271, 10.7125822168258,
> 10.7119832145823, 10.7113816139594, 10.7107780960397, 10.7101732860418,
> 10.7095677728307, 10.7089620284128, 10.7083564497153, 10.7077512971194,
> 11.548875536071, 11.4634458099448, 11.4113675486745, 11.3384424250672,
> 11.2435706626324, 11.1313969585720, 11.0086560681222, 10.8827443523793,
> 10.7598371816865, 10.6440424747848, 10.5382165128003, 10.4442220905656,
> 10.3633207905823, 10.2961499250469, 10.2427320635721, 10.2025802100475,
> 10.1749531325293, 10.1590477762319, 10.1540156426321), sigma = c(0.0947487228789027,
> 0.0947760295260326, 0.0948033853581562, 0.0948307832769866, 0.094858216728106,
> 0.0948856796527442, 0.0949131660004063, 0.0949406718763748, 0.0949681949273155,
> 0.0949957322607503, 0.0950232806230888, 0.095050836445582, 0.0950783958990592,
> 0.0951059550037287, 0.0951335102859937, 0.0951610590705954, 0.0951885984623664,
> 0.0952161256367413, 0.0952436392777666, 0.0952711384472643, 0.0952986226318235,
> 0.0953260918098295, 0.108394172852678, 0.112555919942990, 0.114345649992535,
> 0.115763779372203, 0.116984886895669, 0.118065092089138, 0.119029362771532,
> 0.119887968678076, 0.120638553936562, 0.121278180095107, 0.121810743569063,
> 0.122245010348365, 0.122590801228219, 0.122858869689557, 0.123059216409329,
> 0.123199542683827, 0.123286339009648, 0.123324768295488, 0.123319375423601
> )), .Names = c("age", "country", "mu", "sigma"), row.names = c("580",
> "581", "582", "583", "584", "585", "586", "587", "588", "589",
> "590", "591", "592", "593", "594", "595", "596", "597", "598",
> "599", "600", "601", "602", "603", "604", "605", "606", "607",
> "608", "609", "610", "611", "612", "613", "614", "615", "616",
> "617", "618", "619", "620"), class = "data.frame")
>>
>> result <- lapply(split(musigma.lat.m, musigma.lat.m$country), function(.ctry){
> + # keep all < 5 and only integers over 5
> + subset(.ctry, .ctry$age < 5 | .ctry$age %in% 5:50)
> + })
>>
>> result
> $Bolivia
> age country mu sigma
> 1 0.00000000 Bolivia 11.42168 0.10148719
> 2 0.08333333 Bolivia 11.33625 0.10538375
> 3 0.16666667 Bolivia 11.28417 0.10705943
> 4 0.25000000 Bolivia 11.21125 0.10838720
> 5 0.33333333 Bolivia 11.11637 0.10953050
> ...
> 59 4.83333333 Bolivia 10.49080 0.10671819
> 60 4.91666667 Bolivia 10.48562 0.10653400
> 109 9.00000000 Bolivia 10.43279 0.10180158
> 133 11.00000000 Bolivia 10.33394 0.10160484
> 169 14.00000000 Bolivia 10.24878 0.09946659
> 193 16.00000000 Bolivia 10.20148 0.09694376
> 205 17.00000000 Bolivia 10.16589 0.09573946
>
> $Brazil
> age country mu sigma
> 602 0.00000000 Brazil 11.54888 0.10839417
> 603 0.08333333 Brazil 11.46345 0.11255592
> 604 0.16666667 Brazil 11.41137 0.11434565
> 605 0.25000000 Brazil 11.33844 0.11576378
> ...
> 660 4.83333333 Brazil 10.61799 0.11398118
> 661 4.91666667 Brazil 10.61281 0.11378445
> 710 9.00000000 Brazil 10.55999 0.10872996
> 734 11.00000000 Brazil 10.46113 0.10851983
> 770 14.00000000 Brazil 10.37597 0.10623606
> 794 16.00000000 Brazil 10.32867 0.10354153
>
> -----Original Message-----
> From: jim holtman [mailto:jholtman at gmail.com]
> Sent: 04 November 2009 03:12
> To: Hayes, Daniel
> Cc: r-help at lists.R-project.org
> Subject: Re: [R] creating mulptiple new variables from one data.frame according to columns and rows in that frame
>
> try this:
>
>> x <- read.table(textConnection(" Age(yrs) country mu sigma
> + 1 0.00000000 Bolivia 11.42168 0.1014872
> + 2 0.08333333 Bolivia 11.33625 0.1053837
> + 3 0.16666667 Bolivia 11.28417 0.1070594
> + 4 0.25000000 Bolivia 11.21125 0.1083872
> + 5 0.33333333 Bolivia 11.11637 0.1095305
> + 5.1 5 Bolivia 11.11637 0.1095305
> + 5.2 5.5 Bolivia 11.11637 0.1095305
> + 5.3 6 Bolivia 11.11637 0.1095305
> + 5.4 20 Bolivia 11.11637 0.1095305
> + 5.5 20.1 Bolivia 11.11637 0.1095305
> + 5.6 50 Bolivia 11.11637 0.1095305
> + 602 0.00000000 Brazil 11.54888 0.10839417
> + 603 0.08333333 Brazil 11.46345 0.11255592
> + 604 0.16666667 Brazil 11.41137 0.11434565
> + 605 0.25000000 Brazil 11.33844 0.11576378
> + 606 0.33333333 Brazil 11.24357 0.11698489"), header=TRUE)
>> closeAllConnections()
>> result <- lapply(split(x, x$country), function(.ctry){
> + # keep all < 5 and only integers over 5
> + subset(.ctry, .ctry$Age.yrs. < 5 | .ctry$Age.yrs. %in% 5:50)
> + })
>>
>> result
> $Bolivia
> Age.yrs. country mu sigma
> 1 0.00000000 Bolivia 11.42168 0.1014872
> 2 0.08333333 Bolivia 11.33625 0.1053837
> 3 0.16666667 Bolivia 11.28417 0.1070594
> 4 0.25000000 Bolivia 11.21125 0.1083872
> 5 0.33333333 Bolivia 11.11637 0.1095305
> 5.1 5.00000000 Bolivia 11.11637 0.1095305
> 5.3 6.00000000 Bolivia 11.11637 0.1095305
> 5.4 20.00000000 Bolivia 11.11637 0.1095305
> 5.6 50.00000000 Bolivia 11.11637 0.1095305
>
> $Brazil
> Age.yrs. country mu sigma
> 602 0.00000000 Brazil 11.54888 0.1083942
> 603 0.08333333 Brazil 11.46345 0.1125559
> 604 0.16666667 Brazil 11.41137 0.1143456
> 605 0.25000000 Brazil 11.33844 0.1157638
> 606 0.33333333 Brazil 11.24357 0.1169849
>
>
> On Tue, Nov 3, 2009 at 9:31 AM, Hayes, Daniel <D.J.Hayes at liverpool.ac.uk> wrote:
>> Dear R-helpers,
>>
>> I have a data.frame (bcpe.lat.m) containing 13 countries, ages 0-50yrs per month, and the corresponding mu&sigma (see below).
>>
>> * I would like to limit the age range to include all 12 months for the 1st 5 years and only whole years for all ages thereafter for each of the countries present in the data frame.
>>
>> * I would like to create separate data.frames according to the country the data is from (Bolivia.bcpe.lat.m, brazil.bcpe.lat.m, etc)
>>
>>
>> I have tried using: c(seq(0,5,1/12),seq(5,50,1) ) to select the desired ages but am unsure how to repeat that sequence for consecutive countries.
>> I have tried using: split(bcpe.lat.m, bcpe.lat.m$country) But end up with a string which I am no longer to select the specific ages I want and all the data still remains in one variable
>> Have also looked a 'by', 'apply' and things like 'for (i in 1:13)'
>>
>> Help with either or both steps would be greatly appreciated.
>>
>> Greetings from Formentera,
>> Daniel
>>
>> Age(yrs) country mu sigma
>> 1 0.00000000 Bolivia 11.42168 0.1014872
>> 2 0.08333333 Bolivia 11.33625 0.1053837
>> 3 0.16666667 Bolivia 11.28417 0.1070594
>> 4 0.25000000 Bolivia 11.21125 0.1083872
>> 5 0.33333333 Bolivia 11.11637 0.1095305
>> ...
>> 602 0.00000000 Brazil 11.54888 0.10839417
>> 603 0.08333333 Brazil 11.46345 0.11255592
>> 604 0.16666667 Brazil 11.41137 0.11434565
>> 605 0.25000000 Brazil 11.33844 0.11576378
>> 606 0.33333333 Brazil 11.24357 0.11698489
>> ...
>>
>>
>> [[alternative HTML version deleted]]
>>
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>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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