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RealePrimavera
Obsidian | Level 7

Dear all,

 

I am trying to create customized bins for my unevenly distributed data, skewed on left. I'd like to create more bins on the left side and less on the right side to visualize this distribution.

I have tried proc hpbin but doesnt allow to create uneven bins; alternatively, tried bin function that doesn't work (not sure why). And other numerous solutions proc univariate, proc format etc: nothing worked for what I would like; below is my hypothetical example of bins range that I would like. Thank you very much.

 

Cutoff values
1500
1000
500
250
100
50
30
15
5
3
2
1

 

 

 

 

1 ACCEPTED SOLUTION

Accepted Solutions
PeterClemmensen
Tourmaline | Level 20

How about this?

 

proc format;
   value bin 
      1 <- 5 = "1"
      5 <- 10 = "2"
      10 <- 20 = "3"
      20 <- 30 = "4"
      30 <- 50 = "5"
      50 <- 100 = "6"
      100 <- 200 = "7"
      200 <- 500 = "8"
      500 <- 1000 = "9"
      1000 <- 1500 = "10"
;
run;

data have(drop=i);
    do i=1 to 100;
        x=ceil(rand('uniform')*1500);
        output;
    end;
run;

data want;
    set have;
    id=put(x, bin.);
run;

View solution in original post

7 REPLIES 7
PeterClemmensen
Tourmaline | Level 20

So your bins ranges here are 1500-1000, 1000-500 and so on?

 

Can you show us your PROC FORMAT code? Sounds to me like a job for PROC FORMAT

RealePrimavera
Obsidian | Level 7

Ranges in proc format should roughly look like:

1 <- 5 = "1"
5 <- 10 = "2"
10 <- 20 = "3"
20 <- 30 = "4"
30 <- 50 = "5"
50 <- 100 = "6"
100 <- 200 = "7"
200 <- 500 = "8"
500 <- 1000 = "9"
1000 <- 1500 = "10"

I would ideally like to dump variable x in uneven bins with my choice of cutoff points and then assign those bin levels to another variable 'id', below is example:

 

idxbins of x
…..…..1
…..…..1
…..…..2
…..…..2
…..…..2
…..…..3
…..…..3
…..…..3
…..…..3
…..…..4
…..…..4
…..…..4
…..…..4
…..…..4

 

Thanks,

 

 

PeterClemmensen
Tourmaline | Level 20

How about this?

 

proc format;
   value bin 
      1 <- 5 = "1"
      5 <- 10 = "2"
      10 <- 20 = "3"
      20 <- 30 = "4"
      30 <- 50 = "5"
      50 <- 100 = "6"
      100 <- 200 = "7"
      200 <- 500 = "8"
      500 <- 1000 = "9"
      1000 <- 1500 = "10"
;
run;

data have(drop=i);
    do i=1 to 100;
        x=ceil(rand('uniform')*1500);
        output;
    end;
run;

data want;
    set have;
    id=put(x, bin.);
run;
ballardw
Super User

@PeterClemmensen wrote:

How about this?

 

proc format;
   value bin 
      1 <- 5 = "1"
      5 <- 10 = "2"
      10 <- 20 = "3"
      20 <- 30 = "4"
      30 <- 50 = "5"
      50 <- 100 = "6"
      100 <- 200 = "7"
      200 <- 500 = "8"
      500 <- 1000 = "9"
      1000 <- 1500 = "10"
;
run;

data have(drop=i);
    do i=1 to 100;
        x=ceil(rand('uniform')*1500);
        output;
    end;
run;

data want;
    set have;
    id=put(x, bin.);
run;

To maintain sort order with a format I might suggest using two character values so that the order doesn't become 1, 10, 2 for most purposes.

Or use

id = input(put(x,bin.), 2.);

to create a numeric value.

RealePrimavera
Obsidian | Level 7

Thank you @ballardw

RealePrimavera
Obsidian | Level 7

Thank you very much @PeterClemmensen  for your quick help. 

In fact, I did not even need the second step; there was some mistake in my 'put' code after assigning the bins. The document was not clear to me. 

 

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