awk if-greater-than and replacement under condition - if-statement

Question

I have following data
......
6 4 4 17 154 93 309 0 11930
7 3 2 233 311 0 11936 11932 111874
8 3 1 15 0 11938 11943 211004 11449
9 3 2 55 102 0 11932 11941 111883
10 3 2 197 231 0 11925 11921 111849
11 3 2 160 777 0 11934 11928 111875
......
I hope to replace any values greater than 5000 to 0, from column 4 to column 9. How can I do this work with awk?

Answer 1

To print with lots of spaces like the input, something like this:
awk '{for(i=4;i<=NF;i++)if($i>5000)$i=0; for(i=1;i<=NF;i++)printf "%7d",$i;printf"\n"}' file
Output
6 4 4 17 154 93 309 0 0
7 3 2 233 311 0 0 0 0
8 3 1 15 0 0 0 0 0
9 3 2 55 102 0 0 0 0
10 3 2 197 231 0 0 0 0
11 3 2 160 777 0 0 0 0
For scrunched up together (TM) output, you can use this:
awk '{for(i=4;i<=NF;i++)if($i>5000)$i=0}1' file
6 4 4 17 154 93 309 0 0
7 3 2 233 311 0 0 0 0
8 3 1 15 0 0 0 0 0
9 3 2 55 102 0 0 0 0
10 3 2 197 231 0 0 0 0
11 3 2 160 777 0 0 0 0

Answer 2

An alternative approach (requires gawk4+):
{
patsplit($0, a, "[0-9]+", s)
printf s[0]
for (i=1; i<=length(a); i++){
if(i>4 && a[i]>5000) {
l=length(a[i])
a[i]=0
}
else l=0
printf "%"l"s%s", a[i], s[i]
}
printf "\n"
}
It is more flexible when the spacing would vary, as opposed to the example data. It might also be faster than the accepted answer, in case the number of fields is way bigger than 9.