更新的基准

所以我合成了随机分布的100mn个整数

之间的

0^0 - 1 

and

8^8 - 1

代码生成器

mawk2 '
BEGIN {
     __=_=((_+=_^=_<_)+(__=_*_*_))^(___=__)
     srand()
     ___^=___
     do  { 
           print int(rand()*___) 
  
     } while(--_)  }' | pvE9 > test_large_int_100mil_001.txt

     out9:  795MiB 0:00:11 [69.0MiB/s] [69.0MiB/s] [ <=> ]

  f='test_large_int_100mil_001.txt'
  wc5 < "${f}"

    rows = 100000000. | UTF8 chars = 833771780. | bytes = 833771780.

最后一位的奇/偶分布

Odd  49,992,332
Even 50,007,668

AWK -最快的，有很大的优势(可能C更快，我不知道)

in0:  795MiB 0:00:07 [ 103MiB/s] [ 103MiB/s] [============>] 100%            
( pvE 0.1 in0 < "${f}" | mawk2 '{ _+=$__ } END { print _ }'; )  

 7.64s user 0.35s system 103% cpu 7.727 total
     1  838885279378716

Perl -相当不错

 in0:  795MiB 0:00:10 [77.6MiB/s] [77.6MiB/s] [==============>] 100%            
( pvE 0.1 in0 < "${f}" | perl -lne '$x += $_; END { print $x; }'; )  
 
10.16s user 0.37s system 102% cpu 10.268 total

     1  838885279378716

Python3——稍微落后于Perl

 in0:  795MiB 0:00:11 [71.5MiB/s] [71.5MiB/s] [===========>] 100%            
( pvE 0.1 in0 < "${f}" | python3 -c ; )  

 11.00s user 0.43s system 102% cpu 11.140 total
     1  838885279378716

RUBY -不错

 in0:  795MiB 0:00:13 [61.0MiB/s] [61.0MiB/s] [===========>] 100%            
( pvE 0.1 in0 < "${f}" | ruby -e 'puts ARGF.map(&:to_i).inject(&:+)'; )  
15.30s user 0.70s system 101% cpu 15.757 total

     1  838885279378716

JQ -慢

 in0:  795MiB 0:00:25 [31.1MiB/s] [31.1MiB/s] [========>] 100%            
( pvE 0.1 in0 < "${f}" | jq -s 'add'; )  

 36.95s user 1.09s system 100% cpu 37.840 total

     1  838885279378716

DC

- ( had to kill it after no response in minutes)

给红宝石爱好者

ruby -e "puts ARGF.map(&:to_i).inject(&:+)" numbers.txt

为了完整起见，这里还有一个R解

seq 1 10 | R -q -e "f <- file('stdin'); open(f); cat(sum(as.numeric(readLines(f))))"

《球拍》中的一句俏皮话:

racket -e '(define (g) (define i (read)) (if (eof-object? i) empty (cons i (g)))) (foldr + 0 (g))' < numlist.txt

sed 's/^/.+/' infile | bc | tail -1

dc -f infile -e '[+z1<r]srz1<rp'

注意，带负号前缀的负数应该转换为dc，因为它使用_ prefix而不是- prefix。例如，通过tr '-' '_' | dc -f- -e '…'。

编辑:由于这个答案获得了很多“晦涩难懂”的投票，下面是一个详细的解释:

表达式[+z1<r]srz1<rp的作用如下:

[   interpret everything to the next ] as a string
  +   push two values off the stack, add them and push the result
  z   push the current stack depth
  1   push one
  <r  pop two values and execute register r if the original top-of-stack (1)
      is smaller
]   end of the string, will push the whole thing to the stack
sr  pop a value (the string above) and store it in register r
z   push the current stack depth again
1   push 1
<r  pop two values and execute register r if the original top-of-stack (1)
    is smaller
p   print the current top-of-stack

伪代码:

定义"add_top_of_stack"为: 从堆栈中删除顶部的两个值，并将结果添加回来 如果堆栈有两个或两个以上的值，递归地运行"add_top_of_stack" 如果堆栈有两个或两个以上的值，执行"add_top_of_stack" 打印结果，现在堆栈中只剩下一项

为了真正理解dc的简单和强大，这里有一个工作的Python脚本，它实现了dc的一些命令，并执行上述命令的Python版本:

### Implement some commands from dc
registers = {'r': None}
stack = []
def add():
    stack.append(stack.pop() + stack.pop())
def z():
    stack.append(len(stack))
def less(reg):
    if stack.pop() < stack.pop():
        registers[reg]()
def store(reg):
    registers[reg] = stack.pop()
def p():
    print stack[-1]

### Python version of the dc command above

# The equivalent to -f: read a file and push every line to the stack
import fileinput
for line in fileinput.input():
    stack.append(int(line.strip()))

def cmd():
    add()
    z()
    stack.append(1)
    less('r')

stack.append(cmd)
store('r')
z()
stack.append(1)
less('r')
p()