Dear
I try to create a new subset from my dataframe.
My dataframe's name is m1.
"Classification Description" column has 15 different factors.
The following code is used creating a subset for 1 factor.
m2<-m1[m1$`Classification Description` == levels(m1$`Classification
Description`)[1],]
My aim is to create a subset with 4 different factors. For example,
levels(m1$`Classification Description`)[1]
levels(m1$`Classification Description`)[15]
levels(m1$`Classification Description`)[2]
levels(m1$`Classification Description`)[4]
I try to following code but it didnt work
m2<-m1[m1$`Classification Description` == levels(m1$`Classification
Description`)[c(1,15,2,4],]
How can I solve This Problem ?
Example from my dataframe
`Record Date` `Classification Description` `Current Month
Budget Amount`
<date> <fct>
<dbl>
1 2019-06-30 Total On-Budget and Off-Budget Results:
NA
2 2019-06-30 Off-Budget Surplus (+) or Deficit (-)
41998597035.
3 2019-06-30 Total Outlays
342428650968.
4 2019-06-30 By Other Means
51648504883.
5 2019-06-30 On-Budget Outlays
292169836521.
6 2019-06-30 Off-Budget Outlays
50258814447.
7 2019-06-30 Total Receipts
333952332514.
8 2019-06-30 On-Budget Surplus (+) or Deficit (-)
-50474915489.
9 2019-06-30 Off-Budget Receipts
92257411482
10 2019-06-30 Total On-Budget and Off-Budget Financing
8476318454
--
*Sayg?lar?mla*
Engin YILMAZ
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Hello, Try %in% instead of == in: m2<-m1[m1$`Classification Description` == levels(m1$`Classification Description`)[c(1,15,2,4],] Hope this helps, Rui Barradas ?s 14:57 de 29/07/2020, Engin Y?lmaz escreveu:> Dear > > I try to create a new subset from my dataframe. > My dataframe's name is m1. > "Classification Description" column has 15 different factors. > The following code is used creating a subset for 1 factor. > m2<-m1[m1$`Classification Description` == levels(m1$`Classification > Description`)[1],] > > My aim is to create a subset with 4 different factors. For example, > levels(m1$`Classification Description`)[1] > levels(m1$`Classification Description`)[15] > levels(m1$`Classification Description`)[2] > levels(m1$`Classification Description`)[4] > > I try to following code but it didnt work > > m2<-m1[m1$`Classification Description` == levels(m1$`Classification > Description`)[c(1,15,2,4],] > > How can I solve This Problem ? > > Example from my dataframe > > `Record Date` `Classification Description` `Current Month > Budget Amount` > <date> <fct> > <dbl> > 1 2019-06-30 Total On-Budget and Off-Budget Results: > NA > 2 2019-06-30 Off-Budget Surplus (+) or Deficit (-) > 41998597035. > 3 2019-06-30 Total Outlays > 342428650968. > 4 2019-06-30 By Other Means > 51648504883. > 5 2019-06-30 On-Budget Outlays > 292169836521. > 6 2019-06-30 Off-Budget Outlays > 50258814447. > 7 2019-06-30 Total Receipts > 333952332514. > 8 2019-06-30 On-Budget Surplus (+) or Deficit (-) > -50474915489. > 9 2019-06-30 Off-Budget Receipts > 92257411482 > 10 2019-06-30 Total On-Budget and Off-Budget Financing > 8476318454 > > >-- Este e-mail foi verificado em termos de v?rus pelo software antiv?rus Avast. https://www.avast.com/antivirus
Dear Engin, On 2020-07-29 16:57 +0300, Engin Y?lmaz wrote:> Dear > > I try to create a new subset from my dataframe. > My dataframe's name is m1. > "Classification Description" column has 15 different factors. > The following code is used creating a subset for 1 factor. > m2<-m1[m1$`Classification Description` == levels(m1$`Classification > Description`)[1],] > > My aim is to create a subset with 4 different factors. For example, > levels(m1$`Classification Description`)[1] > levels(m1$`Classification Description`)[15] > levels(m1$`Classification Description`)[2] > levels(m1$`Classification Description`)[4] > > I try to following code but it didnt work > > m2<-m1[m1$`Classification Description` == levels(m1$`Classification > Description`)[c(1,15,2,4],]You're almost correct, you just need to use match instead of ==: m1[m1$`Classification Description` %in% levels(m1$`Classification Description`)[c(1, 15, 2, 4)],] Read more about it at ?match (?`%in%`). Best, Rasmus -------------- next part -------------- A non-text attachment was scrubbed... Name: signature.asc Type: application/pgp-signature Size: 833 bytes Desc: not available URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20200729/dc629e17/attachment.sig>
I solve this as follows m2 <- subset(m1,`Classification Description`=="Borrowing from the Public" | `Classification Description`=="By Other Means" | `Classification Description`=="Total Surplus (+) or Deficit (-)") sincerely Engin YILMAZ Engin Y?lmaz <ispanyolcom at gmail.com>, 29 Tem 2020 ?ar, 16:57 tarihinde ?unu yazd?:> Dear > > I try to create a new subset from my dataframe. > My dataframe's name is m1. > "Classification Description" column has 15 different factors. > The following code is used creating a subset for 1 factor. > m2<-m1[m1$`Classification Description` == levels(m1$`Classification > Description`)[1],] > > My aim is to create a subset with 4 different factors. For example, > levels(m1$`Classification Description`)[1] > levels(m1$`Classification Description`)[15] > levels(m1$`Classification Description`)[2] > levels(m1$`Classification Description`)[4] > > I try to following code but it didnt work > > m2<-m1[m1$`Classification Description` == levels(m1$`Classification > Description`)[c(1,15,2,4],] > > How can I solve This Problem ? > > Example from my dataframe > > `Record Date` `Classification Description` `Current Month > Budget Amount` > <date> <fct> > <dbl> > 1 2019-06-30 Total On-Budget and Off-Budget Results: > NA > 2 2019-06-30 Off-Budget Surplus (+) or Deficit (-) > 41998597035. > 3 2019-06-30 Total Outlays > 342428650968. > 4 2019-06-30 By Other Means > 51648504883. > 5 2019-06-30 On-Budget Outlays > 292169836521. > 6 2019-06-30 Off-Budget Outlays > 50258814447. > 7 2019-06-30 Total Receipts > 333952332514. > 8 2019-06-30 On-Budget Surplus (+) or Deficit (-) > -50474915489. > 9 2019-06-30 Off-Budget Receipts > 92257411482 > 10 2019-06-30 Total On-Budget and Off-Budget Financing > 8476318454 > > > > -- > *Sayg?lar?mla* > Engin YILMAZ >-- *Sayg?lar?mla* Engin YILMAZ [[alternative HTML version deleted]]