大于/小于所有行的awk sed过滤器值

Question

is there a way to construct a filter in awk (or something similar) that for a given file, say:

0.99,0.98,1.1,0.85,0.92
0.76,1.4,0.99,0.99,0.82
1.0,1.45,0.78,0.91,0.95

would replace any record in a line that is greater than 1.0 with 1.0?

Answer 1

Here is something you can do with awk

awk -F, '{for(i=1;i<=NF;i++) if($i>1) {$i="replacement"}}1' OFS=, file

Test:

$ cat file
0.99,0.98,1.1,0.85,0.92
0.76,1.4,0.99,0.99,0.82
1.0,1.45,0.78,0.91,0.95

$ awk -F, '{for(i=1;i<=NF;i++) if($i>1) {$i="replacement"}}1' OFS=, file
0.99,0.98,replacement,0.85,0.92
0.76,replacement,0.99,0.99,0.82
1.0,replacement,0.78,0.91,0.95

Answer 2

Here's a sed solution:

sed -e 's/[1-9][0-9]*\.[0-9]*/1.0/g' in-file > out-file

The pattern [1-9][0-9]*\\.[0-9]* simply matches any sequence that begins with a digit greater than 0 , followed by zero or more digits, followed by the decimal point, followed by additional digits. If you want an in-place replacement, you can use the -i option:

sed -i -e 's/[1-9][0-9]*\.[0-9]*/1.0/g' in-file