Hello, I have a scale and I am trying to create a dummy variable from one of the socio-demographics. The question is what are the sources of your income and I let to pick multiple choices among "help from family", "part-time job", "full-time job" and "scholarship". The dummy I want to create is for measuring financial independence. if any of these response includes "help from family" I want to accept it 0 otherwise 1.
when I use the ifelse function the output was not like what I want;
data$gelkaydummy
I think I see what is happening.
You have a series of answers, one of them being "Help from family." Except sometimes it doesn't get coded properly... But you may be running into an issue with text formatting.
Will the text always be "Help from family" or could it also be "help from family"? You listed both in your question; however, the way your ifelse statement is generated is only searching for a single version of this, in your case the capitalized version.
Here is what I think you can do:
1) Check for unique responses to ensure everything is properly parsed.
unique(data$gelkay)
ought to return 1 instance of each of your categories; if it returns two or more, you have some recoding to do. R and string matching by default see "H" and "h" as different characters.
2) If there is recoding to do, you have some options to pursue. Is it only capitalized letters that are affecting your unique values? You could always convert gelkay to all lowercase.
data$gelkay %
mutate(
gelkay = stringr::str_to_lower(gelkay),
financial_independence = case_when(
str_detect(gelkay,"help from family") ~ 0,
TRUE ~ 1
)
)
Let me know if these work for you. Or, if you are stuck and can't figure out how to fix any issues you encounter should there be any in the unique(), I can help you address those as well.
To my knowledge, R is creating dummy variables automatically. Just check the type of variable in R if it is a factor, then there is no need to create dummy variable . However, if the the variable is not factor convert it using as.factor() built-in function, then put the variable in model you will get the co-efficients for each category.
As I see, some values contain several statements in one string. So, instead of data$gelkay == "Help from family" condition you should use grepl("[Hh]elp from family", data$gelkay).
Atiq Ur Rehman , factors look like strings, but yes, they are actually numeric. Converting using as.numeric() could hypothetically work; however, if they have several potential responses and only want to recode as 0/1 based on one of the options, they would still have to go through and detect the proper rows to attach a 0 while filling in all others as 1, as their column would become 0,1,2,3...
They may be able to use other functions in the purrr package like lump(), but I think that is potentially going a bit overboard if they only want to track a single criteria.
Andrew Vonhandorf Thanks for your answer.
Last night I applied a loop. I divided the response in another data frame after that I coded each response as numeric then I used ifelse function.
Probably it is not the easiest way because when I divided the responses there is space at the start of some values so I need to identify them as extra work.
But it works as how I want.
I am also going to try your advice and let you know about the process.
Regards
gelkay
Alexandr Pozharskiy thanks for that trick. All of my data based on string and I have the problem for capital or lower diffecenre for the same response
Ilhan Albulut Glad to hear you have some progress in terms of it working.
I don't know if you want to do this, but it may be a good idea now that you have a working product to simplify your code. I'm attaching a small .R script that contains an example that I think replicates what you are doing based on what I can tell. Using mutate_at, it will trim the white space (as you mentioned you needed), encode the variables, then create an additional column to determine financial independence based on the value of 1 being present in any of the encoded variables. I purposely use .vars in multiple ways to show how you can approach this through slightly different means, should you need to scale your work. However, if you have several additional columns, you may have to change the financial independence classification to something that is more generalized; maybe using apply or map_lgl.
Ilhan Albulut, try my Aile_f function with apply, as in my example. If it doesn't solve your question, I can help.
library(stringr) # --- You need this library
Aile_f = function(x) {
if(sum(str_detect(toupper(x), "AILE"))) AILE_V = 0 else AILE_V = 1
return(AILE_V)
}
# ---------Reading data (change this to read your data object):
dataobject = read.table(stdin(), header = FALSE, sep = " ")
[1] "ogrenci burs veya kredisi"
[2] "ogrenci burs veya kredisi, Aile destegi"
[3] "ogrenci burs veya kredisi, Yari zamanli calisma"
[4] "ogrenci burs veya kredisi, Yari zamanli calisma"
[5] "ogrenci burs veya kredisi"
[6] "Aile destegi"
[7] "ogrenci burs veya kredisi, Aile destegi"
[8] "Tam zamanli calisma"
[9] "ogrenci burs veya kredisi, Aile destegi"
[10] "ogrenci burs veya kredisi"
dataobject
data_with_dummy = cbind(dataobject, Dummy = apply(dataobject, 1, Aile_f))
data_with_dummy
V1 V2 Dummy
1 [1] ogrenci burs veya kredisi 1
2 [2] ogrenci burs veya kredisi, Aile destegi 0
3 [3] ogrenci burs veya kredisi, Yari zamanli calisma 1
4 [4] ogrenci burs veya kredisi, Yari zamanli calisma 1
5 [5] ogrenci burs veya kredisi 1
6 [6] Aile destegi 0
7 [7] ogrenci burs veya kredisi, Aile destegi 0
8 [8] Tam zamanli calisma 1
9 [9] ogrenci burs veya kredisi, Aile destegi 0
10 [10] ogrenci burs veya kredisi 1
>
good luck
Andrew Vonhandorf Thanks for the script. It works and fits really well. I will also consider your comments.
Jorge Ortiz Pinilla ,
Thanks for your comments and the function.
I applied your function but the output was not similar to yours
[1] 1 0 0 1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 1 0
[21] 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0
[41] 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0
[61] 0 0 0 1 0 1 0 0 1 1 1 0 0 0 1 0 0 0 0 1
[81] 0 0 0 0 0 0 1 0 1 0 1 0 1 1 0 1 0 1 1 1
[101] 0 0 0 0 0 0 0 1 0 0 1 0 0 1 1 1 1 1 0 1
[121] 1 0 1 1 0 1 0 0 1 0 0 0 0 0 0 0 0 1 1 1
[141] 1 1 0 0 0 0 0 0 0 0 1 1 0 1 0 1 0 0 0 0
[161] 0 1 0 0 1 0 0 1 0 1 0 0 0 0 0 1 1 1 1 0
[181] 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0
I tried to make changes to it but I couldn't manage it.
Would you please help me to solve it? Well, I have already a working function but I'd like to learn more and deepen my knowledge in R.
If you please also suggest some sources as well I'd be really happy.
Regards
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