python字符串格式化固定宽度

问题描述

我想把一些浮点数放到一个固定宽度的表中。也就是说，我最多需要12个字符。我想要使​​用至少10个小数（如果有的话）;然而，如果10位小数使其占用超过12个字符，则循环。我原来的想法是尝试像这样

 ＃我只想要12个字符总计
 
 num1 = 0.04154721841 
 num2 = 10.04154721841 
 
＃不是我想要的
 print{：< 12.11g}。format（（num1））
＃想要
打印{：< 12.10f}。format（（num1））
 
＃不是我想要的
打印{：< 12.10f}。格式（（num2））
＃我要
打印{：  

必须有一种方法可以实现这一点，而不必编写一个函数来检查每个数字并根据上述条件进行格式化。我错过了什么？

解决方案

我不确定这是你在找什么，因为它不是完全用格式字符串，但是，当事物变得太长时，您可以使用字符串切片来消除尾随的字符：

$ $ p $ $ $ $ $ $ num1 = 0.04154721841
num2 = 10.04154721841
num3 = 1002.04154721841

print{0：< 12.11g}。format（num1）[：12]
print格式（num2）[：12]
print{0：< 12.11g}。格式（num3）[：12]


$ p $输出：

$ $ $ $ $ $ $ 0.0415472184
10.041547218
1002.0415472

除此之外，我会说你应该只写一个函数，虽然我不是 str.format 东西的专家，所以我可能会遗漏一些东西。

I want to put a bunch of floating point numbers into a fixed-width table. That is, I want a maximum of 12 characters used. I want a minimum of 10 decimals used (if available); however, if 10 decimals makes it take up more than 12 characters, then round. My original thought was to try something like this

# I only want 12 characters used total

num1 = 0.04154721841
num2 = 10.04154721841

# not what I want
print "{:<12.11g}".format((num1))
# what I want
print "{:<12.10f}".format((num1))

# not what I want
print "{:<12.10f}".format((num2))
# what I want
print "{:<12.11g}".format((num2))

There has to be a way to achieve this without writing a function to check every number and give formatting based on the above conditions. What am I missing?

解决方案

I'm not sure this is what you are looking for, since it's not accomplished entirely with the format string, however, you could just use string slicing to lop-off the trailing chars when things get too long:

num1 = 0.04154721841
num2 = 10.04154721841
num3 = 1002.04154721841

print "{0:<12.11g}".format(num1)[:12]
print "{0:<12.11g}".format(num2)[:12]
print "{0:<12.11g}".format(num3)[:12]

outputs:

0.0415472184
10.041547218
1002.0415472

Beyond that, I'd say you should just write a function, though I'm not an expert on the str.format stuff, so I may be missing something.

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