[英]Bash Awk to print number of letters in each word
嘿所有,所以我正在編寫一個bash腳本,它將采用一行文件,例如:
Hello this is my test sentence.
並計算每個單詞中有多少個字母並生成輸出,例如:
5 4 2 2 4 4
這就是我寫的:
#!/bin/bash
awk 'BEGIN {k=1}
{
for(i=1; i<=NF; i++){
stuff[k]=length($i)
printf("%d ", stuff[k])
k++
}
}
END {
printf("%d ", stuff[k])
printf("\n")
}'
它給了我輸出:
5 4 2 2 4 4
它無法識別句子的最后一個單詞中有多少個字母。 相反,它再次使用倒數第二個數字。 我哪里錯了?
不需要awk:
echo Hello this is my test sentence. | {
read -a words
for ((i=0 ; i<${#words[@]}; i++)) ; do
words[i]=${#words[i]}
done
echo "${words[@]}"
}
僅使用bash:
[ ~]$ str="Hello this is my test sentence"
[ ~]$ for word in $str; do echo -n "${#word} "; done; echo ""
5 4 2 2 4 8
bash數組的另一個解決方案:
[ ~]$ echo $str|(read -a words; for word in "${words[@]}"; do echo -n "${#word} "; done; echo "")
5 4 2 2 4 8
假設通過“單詞”表示由空格分隔的文本,這將按字面count how many letter there are in each word
所要求的count how many letter there are in each word
:
$ cat file
Hello this is my test sentence.
and here is another sentence
$ awk '{for (i=1;i<=NF;i++) $i=gsub(/[[:alpha:]]/,"",$i)}1' file
5 4 2 2 4 8
3 4 2 7 8
如果你想計算所有字符,而不僅僅是字母,那就是:
$ awk '{for (i=1;i<=NF;i++) $i=length($i)}1' file
5 4 2 2 4 9
3 4 2 7 8
$ echo Hello this is my test sentence. | awk '{for (i=1;i<=NF;i++) {printf " " length($i)}; print "" }'
5 4 2 2 4 9
或者,將句子放在一個名為sentence
的文件中:
$ awk '{for (i=1;i<=NF;i++) {printf " " length($i)}; print "" }' sentence
5 4 2 2 4 9
如果我們想要刪除前導空格:
$ awk '{for (i=1;i<=NF;i++) {printf "%s%s",length($i),(i==NF?"\n":FS)} }' sentence
5 4 2 2 4 9
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