[R] Problem with filling dataframe's column

[Rui Barradas]

Às 23:13 de 12/06/2023, javad bayat escreveu:
> Dear Rui;
> Many thanks for the email. I tried your codes and found that the length of
> the "Values" and "Names" vectors must be equal, otherwise the results will
> not be useful.
> For some of the characters in the Layer column that I do not need to be
> filled in the LU column, I used "NA".
> But I need to delete some of the rows from the table as they are useless
> for me. I tried this code to delete entire rows of the dataframe which
> contained these three value in the Layer column: It gave me the following
> error.
> 
>> data3 = data2[-grep(c("_esmdes","_Des Section","0"), data2$Layer),]
>       Warning message:
>        In grep(c("_esmdes", "_Des Section", "0"), data2$Layer) :
>        argument 'pattern' has length > 1 and only the first element will be
> used
> 
>> data3 = data2[!grepl(c("_esmdes","_Des Section","0"), data2$Layer),]
>      Warning message:
>      In grepl(c("_esmdes", "_Des Section", "0"), data2$Layer) :
>      argument 'pattern' has length > 1 and only the first element will be
> used
> 
> How can I do this?
> Sincerely
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> On Sun, Jun 11, 2023 at 5:03 PM Rui Barradas <ruipbarradas using sapo.pt> wrote:
> 
>> Às 13:18 de 11/06/2023, Rui Barradas escreveu:
>>> Às 22:54 de 11/06/2023, javad bayat escreveu:
>>>> Dear Rui;
>>>> Many thanks for your email. I used one of your codes,
>>>> "data2$LU[which(data2$Layer == "Level 12")] <- "Park"", and it works
>>>> correctly for me.
>>>> Actually I need to expand the codes so as to consider all "Levels" in
>> the
>>>> "Layer" column. There are more than hundred levels in the Layer column.
>>>> If I use your provided code, I have to write it hundred of time as
>> below:
>>>> data2$LU[which(data2$Layer == "Level 1")] <- "Park";
>>>> data2$LU[which(data2$Layer == "Level 2")] <- "Agri";
>>>> ...
>>>> ...
>>>> ...
>>>> .
>>>> Is there any other way to expand the code in order to consider all of
>> the
>>>> levels simultaneously? Like the below code:
>>>> data2$LU[which(data2$Layer == c("Level 1","Level 2", "Level 3", ...))]
>> <-
>>>> c("Park", "Agri", "GS", ...)
>>>>
>>>>
>>>> Sincerely
>>>>
>>>>
>>>>
>>>>
>>>> On Sun, Jun 11, 2023 at 1:43 PM Rui Barradas <ruipbarradas using sapo.pt>
>>>> wrote:
>>>>
>>>>> Às 21:05 de 11/06/2023, javad bayat escreveu:
>>>>>> Dear R users;
>>>>>> I am trying to fill a column based on a specific value in another
>>>>>> column
>>>>> of
>>>>>> a dataframe, but it seems there is a problem with the codes!
>>>>>> The "Layer" and the "LU" are two different columns of the dataframe.
>>>>>> How can I fix this?
>>>>>> Sincerely
>>>>>>
>>>>>>
>>>>>> for (i in 1:nrow(data2$Layer)){
>>>>>>              if (data2$Layer == "Level 12") {
>>>>>>                  data2$LU == "Park"
>>>>>>                  }
>>>>>>              }
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>> Hello,
>>>>>
>>>>> There are two bugs in your code,
>>>>>
>>>>> 1) the index i is not used in the loop
>>>>> 2) the assignment operator is `<-`, not `==`
>>>>>
>>>>>
>>>>> Here is the loop corrected.
>>>>>
>>>>> for (i in 1:nrow(data2$Layer)){
>>>>>      if (data2$Layer[i] == "Level 12") {
>>>>>        data2$LU[i] <- "Park"
>>>>>      }
>>>>> }
>>>>>
>>>>>
>>>>>
>>>>> But R is a vectorized language, the following two ways are the idiomac
>>>>> ways of doing what you want to do.
>>>>>
>>>>>
>>>>>
>>>>> i <- data2$Layer == "Level 12"
>>>>> data2$LU[i] <- "Park"
>>>>>
>>>>> # equivalent one-liner
>>>>> data2$LU[data2$Layer == "Level 12"] <- "Park"
>>>>>
>>>>>
>>>>>
>>>>> If there are NA's in data2$Layer it's probably safer to use ?which() in
>>>>> the logical index, to have a numeric one.
>>>>>
>>>>>
>>>>>
>>>>> i <- which(data2$Layer == "Level 12")
>>>>> data2$LU[i] <- "Park"
>>>>>
>>>>> # equivalent one-liner
>>>>> data2$LU[which(data2$Layer == "Level 12")] <- "Park"
>>>>>
>>>>>
>>>>> Hope this helps,
>>>>>
>>>>> Rui Barradas
>>>>>
>>>>
>>>>
>>> Hello,
>>>
>>> You don't need to repeat the same instruction 100+ times, there is a way
>>> of assigning all new LU values at the same time with match().
>>> This assumes that you have the new values in a vector.
>>
>> Sorry, this is not clear. I mean
>>
>>
>> This assumes that you have the new values in a vector, the vector Names
>> below. The vector of values to be matched is created from the data.
>>
>>
>> Rui Barradas
>>
>>>
>>>
>>> Values <- sort(unique(data2$Layer))
>>> Names <- c("Park", "Agri", "GS")
>>>
>>> i <- match(data2$Layer, Values)
>>> data2$LU <- Names[i]
>>>
>>>
>>> Hope this helps,
>>>
>>> Rui Barradas
>>>
>>> ______________________________________________
>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> 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.
>>
>>
> 
Hello,

Please cc the r-help list, R-Help is threaded and this can in the future 
be helpful to others.

You can combine several patters like this:


pat <- c("_esmdes|_Des Section|0")
grep(pat, data2$Layer)

or, programatically,


pat <- paste(c("_esmdes","_Des Section","0"), collapse = "|")


Hope this helps,

Rui Barradas