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printf:如何对齐小数点上的数字

[英]printf: how to align numbers on decimal point

consider having a set of numbers: 考虑有一组数字:

my @array =   (
   1.788139343e-007, 0.0547055073198, -0.703213036125,
   -0.583665880391, -1.41198285298, +0.171879081676,
   -0.58966025098, -86.0627173425, -0.84449797709,
 3.49876623321e-005, 3.02660429162, -0.256948695361);

I would like to have the decimal point aligned at the n-th column of a total width of m (maybe n=6 and m=25) 我想将小数点对齐在总宽度为m的第n列(可能n = 6和m = 25)

If using %f i get nicely aligned numbers, but the numbers that require an scientific notation are being destroyed. 如果使用%f我得到了很好的对齐数字,但是需要科学记数法的数字正在被破坏。 usging %g interprets the precision argument after the dot as an absolute precision that results in different decimals after the decimal point. usging %g将点后面的precision参数解释为绝对精度,导致小数点后面的小数点不同。 And since the most numbers are in the range (-10, 10) I do not want to take the scientific notation %e 由于大多数数字在(-10,10)范围内,我不想采用科学记数法%e

Are there any flags or format attributes (or combination of) that I overlooked? 是否有我忽略的标志或格式属性(或组合)?

the expected outcome would be: 预期的结果将是:

foreach my $f (@array){
  printf("[%+25.12g]$/", $f);
}


[   +1.788139343e-007     ]
[   +0.0547055073198      ]
[   -0.703213036125       ]
[   -0.583665880391       ]
[   -1.41198285298        ]
[   +0.0171879081676      ]
[   -0.58966025098        ]
[  -86.0627173425         ]
[   -0.84449797709        ]
[   +3.49876623321e-005   ]
[   +3.02660429162        ]
[   -0.256948695361       ]

or even better 甚至更好

[   +1.7881393430000e-007 ]
[   +0.0547055073198      ]
[   -0.7032130361250      ]
[   -0.5836658803910      ]
[   -1.4119828529800      ]
[   +0.0171879081676      ]
[   -0.5896602509800      ]
[  -86.0627173425000      ]
[   -0.8444979770900      ]
[   +3.4987662332100e-005 ]
[   +3.0266042916200      ]
[   -0.2569486953610      ]

(the question is about Perl but s?printf s format string is rather independent, so I also added C tag) (问题是关于Perl但是s?printf的格式字符串是相当独立的,所以我也添加了C标签)

The [*]printf function allow you to: [*]printf函数允许您:

  • get how much characters have been written, 得到多少字符,
  • add some padding. 添加一些填充。

So if you know how much characters are before the dot ( d = sprintf(buf, "%.0f", ar[i]); ), you can align the dot using ( printf("[%*s %g", 4-d, "", ar[i]); ). 因此,如果您知道点之前有多少字符( d = sprintf(buf, "%.0f", ar[i]); ),您可以使用( printf("[%*s %g", 4-d, "", ar[i]); )。

Then same logic to align the closing bracket: 然后使用相同的逻辑来对齐右括号:

#include <stdio.h>

int main()
{
    double ar[] = {
        1.788139343e-007,   0.0547055073198, -0.703213036125,
       -0.583665880391,    -1.41198285298,    0.171879081676,
       -0.58966025098,    -86.0627173425,    -0.84449797709,
        3.49876623321e-005, 3.02660429162,   -0.256948695361};

    for (int i = 0; i < 12; ++i)
    {
        /* buffer to count how much character are before the dot*/
        char buf[64];

        /* how much before the dot? */
        int d = sprintf(buf, "%+.0lf", ar[i]);

        /* write float with aligned dot and store second padding */
        int e = printf("[%*s %+.15lg", 4-d, "", ar[i]);
        printf("%*s]\n", 25-e, "");

    }
    return 0;
}

Gives: 得到:

[   +1.788139343e-07     ]
[   +0.0547055073198     ]
[   -0.703213036125      ]
[   -0.583665880391      ]
[   -1.41198285298       ]
[   +0.171879081676      ]
[   -0.58966025098       ]
[  -86.0627173425        ]
[   -0.84449797709       ]
[   +3.49876623321e-05   ]
[   +3.02660429162       ]
[   -0.256948695361      ]

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