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bash - awk: find minimum and maximum in column

I'm using awk to deal with a simple .dat file, which contains several lines of data and each line has 4 columns separated by a single space. I want to find the minimum and maximum of the first column.

The data file looks like this:

9 30 8.58939 167.759
9 38 1.3709 164.318
10 30 6.69505 169.529
10 31 7.05698 169.425
11 30 6.03872 169.095
11 31 5.5398 167.902
12 30 3.66257 168.689
12 31 9.6747 167.049
4 30 10.7602 169.611
4 31 8.25869 169.637
5 30 7.08504 170.212
5 31 11.5508 168.409
6 31 5.57599 168.903
6 32 6.37579 168.283
7 30 11.8416 168.538
7 31 -2.70843 167.116
8 30 47.1137 126.085
8 31 4.73017 169.496

The commands I used are as follows.

min=`awk 'BEGIN{a=1000}{if ($1<a) a=$1 fi} END{print a}' mydata.dat`
max=`awk 'BEGIN{a=   0}{if ($1>a) a=$1 fi} END{print a}' mydata.dat`

However, the output is min=10 and max=9.

(The similar commands can return me the right minimum and maximum of the second column.)

Could someone tell me where I was wrong? Thank you!

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Awk guesses the type.

String "10" is less than string "4" because character "1" comes before "4". Force a type conversion, using addition of zero:

min=`awk 'BEGIN{a=1000}{if ($1<0+a) a=$1} END{print a}' mydata.dat`
max=`awk 'BEGIN{a=   0}{if ($1>0+a) a=$1} END{print a}' mydata.dat`

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