bash - 如何用 shell 脚本替换单位值

Question

我有一个格式为的输出

weight:121p height:6f money:5 update:8 read:62b query:132 etc

我希望输出没有像这样的单位

weight:121 height:6 money:5 update:8 read:62 query:132 etc

我试过| tr "p", " "etc 但问题是它删除了输出中每次出现的字母 p 。

我只想删除冒号后的单位。解决此问题的最佳方法是什么？

Answer 1

sed 's/\(:[0-9][0-9]*\)[a-z]/\1/g'

或者如果单位可能超过一个字母

sed 's/\(:[0-9][0-9]*\)[a-z][a-z]*/\1/g'

Answer 2

echo "weight:121p height:6f money:5 update:8 read:62b query:132 etc"|awk '{$1=substr($1,0,length($1)-1);$2=substr($2,0,length($2)-1);$5=substr($5,0,length($5)-1);print}'

        weight:121 height:6 money:5 update:8 read:62 query:132 etc

Answer 3

如果你更喜欢 perl：

echo "weight:121p height:6f money:5 update:8 read:62b query:132" \
| perl -pe 's/(\w+:\d+)\w* /$1 /g'

Answer 4

这会将带有尾随非数字/非点的数字强制转换为数字。

awk '{for (i = 1; i <= NF; i++) {split($i, a, /:/); if (a[2] ~ /^[[:digit:]]/) {$i = a[1] ":" a[2] + 0}}; print}' inputfile