Linux 如何快速求和文件中的所有数字？
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How can I quickly sum all numbers in a file?
提问by Mark Roberts
I have a file which contains several thousand numbers, each on it's own line:
我有一个包含数千个数字的文件，每个数字都在自己的行上：
34
42
11
6
2
99
...
I'm looking to write a script which will print the sum of all numbers in the file. I've got a solution, but it's not very efficient. (It takes several minutes to run.) I'm looking for a more efficient solution. Any suggestions?
我正在寻找编写一个脚本来打印文件中所有数字的总和。我有一个解决方案，但效率不高。（运行需要几分钟。）我正在寻找更有效的解决方案。有什么建议？
采纳答案by brian d foy
For a Perl one-liner, it's basically the same thing as the awksolution in Ayman Hourieh's answer:
对于 Perl one-liner，它与Ayman Hourieh 的回答中的awk解决方案基本相同：
% perl -nle '$sum += $_ } END { print $sum'
If you're curious what Perl one-liners do, you can deparse them:
如果您对 Perl 单行程序的功能感到好奇，您可以解析它们：
% perl -MO=Deparse -nle '$sum += $_ } END { print $sum'
The result is a more verbose version of the program, in a form that no one would ever write on their own:
结果是程序的一个更冗长的版本，以一种没有人会自己编写的形式：
BEGIN { $/ = "\n"; $\ = "\n"; }
LINE: while (defined($_ = <ARGV>)) {
chomp $_;
$sum += $_;
}
sub END {
print $sum;
}
-e syntax OK
Just for giggles, I tried this with a file containing 1,000,000 numbers (in the range 0 - 9,999). On my Mac Pro, it returns virtually instantaneously. That's too bad, because I was hoping using mmapwould be really fast, but it's just the same time:
只是为了傻笑，我用一个包含 1,000,000 个数字（在 0 - 9,999 范围内）的文件进行了尝试。在我的 Mac Pro 上，它几乎立即恢复。这太糟糕了，因为我希望使用mmap会非常快，但它只是在同一时间：
use 5.010;
use File::Map qw(map_file);
map_file my $map, $ARGV[0];
$sum += while $map =~ m/(\d+)/g;
say $sum;
回答by Stefan Kendall
I don't know if you can get a lot better than this, considering you need to read through the whole file.
考虑到您需要通读整个文件，我不知道您是否可以获得比这更好的东西。
$sum = 0;
while(<>){
$sum += $_;
}
print $sum;
回答by Ayman Hourieh
You can use awk:
您可以使用 awk：
awk '{ sum += } END { print sum }' file
回答by frankc
Just to be ridiculous:
只是为了可笑：
cat f | tr "\n" "+" | perl -pne chop | R --vanilla --slave
回答by Paused until further notice.
This is straight Bash:
这是直接的 Bash：
sum=0
while read -r line
do
(( sum += line ))
done < file
echo $sum
回答by DVK
I have not tested this but it should work:
我还没有测试过这个，但它应该可以工作：
cat f | tr "\n" "+" | sed 's/+$/\n/' | bc
You might have to add "\n" to the string before bc (like via echo) if bc doesn't treat EOF and EOL...
如果 bc 不处理 EOF 和 EOL，您可能必须在 bc 之前的字符串中添加“\n”（如通过 echo）...
回答by ghostdog74
sed ':a;N;s/\n/+/;ta' file|bc
回答by lhf
Here's another one-liner
这是另一个单线
( echo 0 ; sed 's/$/ +/' foo ; echo p ) | dc
This assumes the numbers are integers. If you need decimals, try
这假设数字是整数。如果您需要小数，请尝试
( echo 0 2k ; sed 's/$/ +/' foo ; echo p ) | dc
Adjust 2 to the number of decimals needed.
将 2 调整为所需的小数位数。
回答by nickjb
Another for fun
另一个好玩的
sum=0;for i in $(cat file);do sum=$((sum+$i));done;echo $sum
or another bash only
或仅其他 bash
s=0;while read l; do s=$((s+$l));done<file;echo $s
But awk solution is probably best as it's most compact.
但是 awk 解决方案可能是最好的，因为它最紧凑。
回答by ruben2020
Here's another:
这是另一个：
open(FIL, "a.txt");
my $sum = 0;
foreach( <FIL> ) {chomp; $sum += $_;}
close(FIL);
print "Sum = $sum\n";
