[英]Python Format Decimal with a minimum number of Decimal Places
I have some Decimal
instances in Python. 我在Python中有一些Decimal
实例。 I wish to format them such that 我希望将它们格式化
Decimal('1') => '1.00'
Decimal('12.0') => '12.00'
Decimal('314.1') => '314.10'
Decimal('314.151') => '314.151'
hence ensuring that there are always at least two decimal places, possibly more. 因此确保总是至少有两个小数位,可能更多。 While there are no shortage of solutions for rounding to n
decimal places I can find no neat ways of ensuring a lower bound on the number. 虽然不存在舍入到n
个小数位的解决方案,但我找不到确保数字下限的简洁方法。
My current solution is to compute: 我目前的解决方案是计算:
first = '{}'.format(d)
second = '{:.2f}'.format(d)
and take which ever of the two is longer. 并采取两者中的哪一个更长。 However it seems somewhat hackish. 然而,它看起来有点hackish。
If you wish to avoid string issues: 如果您希望避免字符串问题:
if d*100 - int(d*100):
print str(d)
else:
print ".2f" % d
Untested code, but it should work. 未经测试的代码,但它应该工作。
This works like so: 这样工作如下:
d = 12.345 d = 12.345
Times 100: 时代100:
1234.5 1234.5
Minus int(1234.5) 减去int(1234.5)
1234.5 - 1234 = .5 1234.5 - 1234 = .5
.5 != 0 .5!= 0
This means that there are 3 or more decimal places. 这意味着有3个或更多小数位。
print str(12.345) print str(12.345)
Even if you do 12.3405: 即使你做12.3405:
1234.05 - 1234 = .05 1234.05 - 1234 = .05
.05 != 0 .05!= 0
But if you have 12.3: 但是如果你有12.3:
1230 - 1230 = 0 1230 - 1230 = 0
This means to print with %.2f. 这意味着使用%.2f进行打印。
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