使用现有文件:

paste -sd+ infile | bc

使用标准输入:

<cmd> | paste -sd+ | bc

编辑: 用一些 在阅读时需要更加明确的实现

<cmd> | paste -sd+ - | bc

:

-s(串行)-将所有行合并为一行

+ 在这种情况下)

我喜欢选择的答案。但是,它往往比awk慢,因为需要两个工具来完成这项工作。

$ wc -l file
49999998 file

$ time paste -sd+ file | bc
1448700364

real    1m36.960s
user    1m24.515s
sys     0m1.772s

$ time awk '{s+=$1}END{print s}' file
1448700364

real    0m45.476s
user    0m40.756s
sys     0m0.287s

以下命令将添加所有行(awk输出的第一个字段)

awk '{s+=$1} END {print s}' filename

awk '{ sum += $1; }
     END { print sum; }' "$@"

sum < FileWithColumnOfNumbers.txt
sum   FileWithColumnOfNumbers.txt

FWIW:在MacOS X上,您可以使用一行程序:

awk '{ sum += $1; } END { print sum; }' "$@"

【Ghostdog74评论后续】

bash-2.03$ uname -sr
SunOS 5.8

bash-2.03$ perl -le 'print for 1..49999998' > infile

bash-2.03$ wc -l infile
 49999998 infile

bash-2.03$  time paste -sd+ infile | bc
bundling space exceeded on line 1, teletype
Broken Pipe

real    0m0.062s
user    0m0.010s
sys     0m0.010s

bash-2.03$ time nawk '{s+=$1}END{print s}' infile
1249999925000001

real    2m0.042s
user    1m59.220s
sys     0m0.590s
bash-2.03$ time /usr/xpg4/bin/awk '{s+=$1}END{print s}' infile
1249999925000001

real    2m27.260s
user    2m26.230s
sys     0m0.660s

bash-2.03$ time perl -nle'
  $s += $_; END { print $s }
   ' infile
1.249999925e+15

real    1m34.663s
user    1m33.710s
sys     0m0.650s

$ cat n
1
2
3
$ (cat n | tr "\012" "+" ; echo "0") | bc 
6

这个 tr 将所有换行符更改为“+”;然后在最后一个加号后面加上0,然后通过管道传递表达式( 1+2+3+0

或者,如果您可以使用awk或perl,这里有一个perl一行程序:

$perl -nle '$sum += $_ } END { print $sum' n
6

while read -r num; do ((sum += num)); done < inputfile; echo $sum

使用 for

sum=0; for x in `cat <your-file>`; do let sum+=x; done; echo $sum

如果你安装了ruby

cat FileWithColumnOfNumbers.txt | xargs ruby -e "puts ARGV.map(&:to_i).inject(&:+)"

[root@pentest3r ~]# (find / -xdev -size +1024M) | (while read a ; do aa=$(du -sh $a | cut -d "." -f1 ); o=$(( $o+$aa )); done; echo "$o";)