[英]printf: how to align numbers on decimal point
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);
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)
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
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。 由于大多数数字在(-10,10)范围内,我不想采用科学记数法%e
是否有我忽略的标志或格式属性(或组合)?
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 ]
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 ]
(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)
[*]printf
函数允许您:
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]);
)。
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;
}
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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