[R] Getting minimum value of a column according a factor column of a dataframe

@vi@e@gross m@iii@g oii gm@ii@com @vi@e@gross m@iii@g oii gm@ii@com
Thu Aug 25 23:50:45 CEST 2022

The requirements keep being clarified and it would have been very useful to know more in advance.


To be clear. My earlier suggestion was based on JUST wanting the minimum for each unique version of Code. Then you wanted it in the original order so that was handled by carefully making that a factor column in the order you wanted the output. Now the request is to throw back in ALL the columns for as many rows as are deemed minimums.


So, not in any way demeaning the various methods others offer, I suggest you look at database style joins. R has some in the core with names like merge() and other packages such as dplyr have all kinds of variations on a theme.


In my case, you can extend the part of my code that makes this as a second data.frame/tibble:


   Code  minQ

1 41003 0.160

2 88888 0.160

3 41005 0.210

4 41009 0.218

5 41017 0.240


Call it df2 as in:


mydf.min <-

  mydf %>%

  group_by(Code) %>%

  summarize(minQ = min(Q))


You now have the original thing I called mydf that has a column called Code and lots of other columns and you have a smaller one with fewer rows and columns called mydf.min and they share a single common column.


You want to merge these two using whatever kind of join makes sense. Dplyr offers an inner_join(), left_join(), right_join() and full_join() and you can tweak merge() and others to do similar things.


What you seem to want is to find all rows that share both a particular value for Code and at the same time a particular value for minQ in one versus Q in the other. You want to ignore all others.  What gets returned can have all the original columns and perhaps also the minQ column (which can be removed) and if two or more rows in one grouping share exactly the same minimum, may get slightly similar but different lines matched. Is that what you want? Do note matching floating point equally can be dangerous but in this case since the numbers were all just read in, should be fine.


You can play with it but I tried this:


test <- left_join(mydf.min, mydf, by=c("Code", "minQ" = "Q"))


The values returned (and not I added an 88888 category earlier in my data) look like this:


Code  minQ  Y  M  D    N    O

1 41003 0.160 81  1 19 7.17 2.50

2 88888 0.160 81  1 19 7.17 2.50

3 41005 0.210 79  8 17 5.50 7.20

4 41005 0.210 80 10 30 6.84 2.60

5 41005 0.210 80 12 20 6.84 2.40

6 41009 0.218 79  2 21 5.56 4.04

7 41009 0.218 79  5 27 6.40 3.12

8 41017 0.240 79 10 20 5.30 7.10

9 41017 0.240 80  7 30 6.73 2.60


You can see there are three minimum rows for Code 41005 for example and the join keeps them all.


You can trivially remove the minQ column by naming the original as Q or just removing it. You can again reorder things if you wish including by sorting on other columns ascending or descending using other functions/verbs.


As noted, this has to work for you with your larger code set and sometimes complex enough such requirements can be done many ways as it can be as much art as science. I personally would probably write most of the above as one long pipeline looking a bit like:


mydf <-   

  read.table(…) %>%

  mutate(Code=factor(..., levels=unique(...)))


result <-

  mydf %>%

  group_by(Code) %>%

  summarize(...) %>%

  left_join(mydf, by=c("Code", "minQ" = "Q")) %>%



And of course it may be better to use the new R pipe operator if your version is new and so on, filling in whatever details make sense to you. At the end of the pipeline, you might want to use verbs that sort the data as described above.


My guess is you will now tell us about yet another condition suggestions like mine do not fulfill and I will likely then ignore …





From: javad bayat <j.bayat194 using gmail.com> 
Sent: Thursday, August 25, 2022 2:02 PM
To: avi.e.gross using gmail.com
Cc: R-help using r-project.org
Subject: Re: [R] Getting minimum value of a column according a factor column of a dataframe


;Dear all

First of all I appreciate you for the answers you have sent. I did the codes that Rui provided and I got what I wanted.


res <- lapply(split(df1, df1$Code), \(x) x[which.min(x$Q),])
res <- do.call(rbind, res)
i <- order(unique(df1$Code))
res[order(i), ] 


I think I should explain more about my request. I had a large data frame (11059 rows and 16 columns). The Code column represented the stations code, totally the number of stations were 128. At each station I had many measured variables, like Q and N and O, and these variables were measured in different years and days. The days that data were measured were different for each station, and due to this reason I had different rows for stations. For example, station number one (41009) had 158 rows and station number 2 (41011) had 113 rows. Note that the station's codes are not in order format (e.g smallest to largest). 

Back to my request, I wanted to extract the minimum value of the Q for each station (based on the Code column). The problem was that I wanted to have other column values which were measured for this minimum of the Q. 

I hope my explanation was clear enough. As I said before, I used the Rui's codes and I got what I wanted. Although, other solutions provided by others were all correct. 


Regarding my request, unfortunately I faced another problem. I had to extract the maximum of the Q and put it exactly under the minimum of the Q. Something like the below one:










































I extract both min and max according to the codes, and I have 2 dataframes, one for the minimum values and another for the max values. Both dataframe have a Code column which is exactly similar.

Can I extract both min and max simultaneously or I have to combine two dataframes?

I used the rbind and merge function but they did not give the desired results.

> df3 = merge (df1, df2, by = "Code")

The result of this code adds a second dataframe as columns to the first one. I want the first row of the second dataframe put below the first row of the first dataframe and so on. I used a function to do this but it seems it does not work correctly.


> fun2 = function(x,y){
                i = 1
                for(i in x) {
                      if (x[i,1] == y[i,1]){
                          i = i+1
> fun2(df1, df2)































On Thu, Aug 25, 2022 at 9:08 PM <avi.e.gross using gmail.com <mailto:avi.e.gross using gmail.com> > wrote:

Yes, Timothy, the request was not seen by all of us as the same.

Indeed if the request was to show a subset of the original data consisting
of only the rows that were the minimum for each Code and also showed ties,
then the solution is a tad more complex. I would then do something along the
lines of what others showed such as generating another column showing the
minimum for each row and then showing only rows that matched their value in
two columns or whatever was needed.

As noted, keeping the output in a specific order was not initially

Keeping the data in some order is a common enough request but in this
situation, I suspect the order many might want would be the one showing the
minimums in order, not the codes in the original order. 

-----Original Message-----
From: Ebert,Timothy Aaron <tebert using ufl.edu <mailto:tebert using ufl.edu> > 
Sent: Thursday, August 25, 2022 11:59 AM
To: avi.e.gross using gmail.com <mailto:avi.e.gross using gmail.com> 
Cc: R-help using r-project.org <mailto:R-help using r-project.org> 
Subject: RE: [R] Getting minimum value of a column according a factor column
of a dataframe

My assumption (maybe wrong) was that we needed to keep the other variables.
I want to find the values of Y, M, D, N, and O for the minimum value of Q
within each unique value of Code, keeping the data in the original order.
All one need to do is filter Q in the original dataframe by your answer for

Keeping the data in the original order seems unnecessary, but that is what
was asked in a later post.

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Best Regards
Javad Bayat
M.Sc. Environment Engineering
Alternative Mail: bayat194 using yahoo.com <mailto:bayat194 using yahoo.com> 

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