格式化数字，使它们在小数点上对齐

在Python中，我需要格式化数字，使它们在小数点上对齐，如下所示:

最直接的方法是什么?

如果您知道所需的精度(小数点后的数字)，并且您不介意在使用整数时末尾有一些零，您可以使用Python 3.6 (PEP498)中的新f-string:

numbers = [4.8, 49.723, 456.781, -72.18, 5, 13]
for number in numbers:
    print(f'{number:9.4f}')

打印:

我不认为有一种直接的方法可以做到这一点，因为在开始打印它们之前，您需要知道所有数字的小数点的位置。(我刚刚看了一下Caramiriel的链接，以及那个页面上的一些链接，但我找不到任何特别适用于这种情况的链接)。

看起来你需要做一些基于字符串的检查&对列表中的数字进行操作。例如,

def dot_aligned(seq):
    snums = [str(n) for n in seq]
    dots = [s.find('.') for s in snums]
    m = max(dots)
    return [' '*(m - d) + s for s, d in zip(snums, dots)]
nums = [4.8, 49.723, 456.781, -72.18]
for s in dot_aligned(nums):
    print(s)

如果你想处理一个包含一些普通int的float s列表，那么这种方法会变得有点混乱。

def dot_aligned(seq):
    snums = [str(n) for n in seq]
    dots = []
    for s in snums:
        p = s.find('.')
        if p == -1:
            p = len(s)
        dots.append(p)
    m = max(dots)
    return [' '*(m - d) + s for s, d in zip(snums, dots)]
nums = [4.8, 49.723, 456.781, -72.18, 5, 13]
for s in dot_aligned(nums):
    print(s)

正如Mark Ransom在评论中指出的那样，我们可以通过使用.split来简化int s的处理:

def dot_aligned(seq):
    snums = [str(n) for n in seq]
    dots = [len(s.split('.', 1)[0]) for s in snums]
    m = max(dots)
    return [' '*(m - d) + s for s, d in zip(snums, dots)]

Masher在评论中提到，在右侧添加填充可以很有用，这样数字就可以在对齐的列中打印。然而，我们不需要为每个字符串计算填充的大小，我们可以使用str.ljust方法。

def dot_aligned(seq):
    snums = [str(n) for n in seq]
    dots = [len(s.split('.', 1)[0]) for s in snums]
    m = max(dots)
    left_pad = [' '*(m - d) + s for s, d in zip(snums, dots)]
    ms = max(map(len, left_pad))
    return [s.ljust(ms) for s in left_pad]
nums = [4.8, 49.723, 456.781, -72.18, 5, 13, 1.2345] * 3
cols = 4
# Get number of cells in the output grid, using ceiling division
size = len(nums) // -cols * -cols
padded = dot_aligned(nums)
for i in range(0, size, cols):
    print(*padded[i:i+cols])

  4.8     49.723  456.781  -72.18  
  5       13        1.2345   4.8   
 49.723  456.781  -72.18     5     
 13        1.2345   4.8     49.723 
456.781  -72.18     5       13     
  1.2345

如果您不介意后面的零，那么简单的方法是:

numbers = [4.8,49.723,456.781,-72.18,5,13]
for f in numbers:
    print('{:>7.3f}'.format(f))

打印:

如果你想去掉后面的零，你可以使用正则表达式模块中的re.sub方法:

import re
numbers = [4.8,49.723,456.781,-72.18,5,13]
for f in numbers:
    print(re.sub(r'.?0+$','','{:>7.3f}'.format(f)))

打印:

但是，这会给你不同宽度的列。唯一的区别是空格，所以你不能看到它，但如果你把它作为表的一部分，它看起来像这样:

import re
numbers = [4.8,49.723,456.781,-72.18,5,13]
for f in numbers:
    print(re.sub(r'.?0+$','','{:>7.3f}'.format(f)),'|')

打印:

  4.8 |
 49.723 |
456.781 |
-72.18 |
  5 |
 13 |

为了避免这种情况，如果你想让真正的花哨，你可以这样做:

import re
numbers = [4.8,49.723,456.781,-72.18,5,13]
for f in numbers:
    print(re.sub(r'.?0+$',lambda match: ' '*(match.end()-match.start()),'{:>7.3f}'.format(f)),'|')

打印:

希望这对你有帮助!

如果你事先知道你需要的前导空格和十进制数字的个数，就像在其他回复中一样，最简单的方法是

    # python 2 version
    numbers = [4.8, 49.723, 456.781, -72.18, 5, 13, 0.1, .6666, 50000, -40000]
    for number in numbers:
        print '{:16.4f}'.format(number).rstrip('0').rstrip('.')

    # python 3 version
    numbers = [4.8, 49.723, 456.781, -72.18, 5, 13, 0.1, .6666, 50000, -40000]
    for number in numbers:
        print f'{number:16.4f}'.rstrip('0').rstrip('.')

输出:

作为PM 2Ring的答案的替代方案，要动态计算点列的正确位置，您可以使用以下解决方案之一:

# python 3, f-string and .format() mixed version
numbers = [4.8, 49.723, 456.781, -72.18, 5, 13, 0.1, .6666, 50000, -40000]
numbers2string = [str(X) for X in numbers]
numbers_splitted = [X.split(".") for X in numbers2string]
len_max_before = max([len(X[0]) for X in numbers_splitted])
len_max_after = max([len(X[1]) for X in numbers_splitted if len(X) > 1])
len_max_total = len_max_before + len_max_after + 1
for n in numbers:
    numstring = f'{"{0: >#0"}{len_max_total}.{len_max_after}f{"}"}'
    print(numstring.format(n).rstrip('0').rstrip('.'))

# python 3, .format() version
numbers = [4.8, 49.723, 456.781, -72.18, 5, 13, 0.1, .6666, 50000, -40000]
numbers2string = [str(X) for X in numbers]
numbers_splitted = [X.split(".") for X in numbers2string]
len_max_before = max([len(X[0]) for X in numbers_splitted])
len_max_after = max([len(X[1]) for X in numbers_splitted if len(X) > 1])
for number in numbers2string:
    if '.' in number:
        number = number.split('.')
        print("{number[0]:>{len_max_before}}.{number[1]:<{len_max_before}}".format(
            number=number,
            len_max_before=len_max_before,
            len_max_after=len_max_after
        ))
    else:
        print("{number:>{len_max_before}}".format(
            number=number,
            len_max_before=len_max_before
        ))

# python 2 version
numbers = [4.8, 49.723, 456.781, -72.18, 5, 13, 0.1, .6666, 50000, -40000]
numbers2string = [str(X) for X in numbers]
numbers_splitted = [X.split(".") for X in numbers2string]
len_max_before = max([len(X[0]) for X in numbers_splitted])
len_max_after = max([len(X[1]) for X in numbers_splitted if len(X) > 1])
for number in numbers2string:
    if '.' in number:
        number = number.split('.')
        print "{number[0]:>{len_max_before}}.{number[1]:<{len_max_before}}".format(
            number=number,
            len_max_before=len_max_before,
            len_max_after=len_max_after
        )
    else:
        print "{number:>{len_max_before}}".format(
            number=number,
            len_max_before=len_max_before
        )

# python 3 f-string version
numbers = [4.8, 49.723, 456.781, -72.18, 5, 13, 0.1, .6666, 50000, -40000]
numbers2string = [str(X) for X in numbers]
numbers_splitted = [X.split(".") for X in numbers2string]
len_max_before = max([len(X[0]) for X in numbers_splitted])
len_max_after = max([len(X[1]) for X in numbers_splitted if len(X) > 1])
for number in numbers2string:
    if '.' in number:
        number = number.split('.')
        numstring = f"{number[0]:>{len_max_before}}.{number[1]:<{len_max_after}}"
    else:
        numstring = f"{number:>{len_max_before}}"
    print(numstring)

输出:

使用Python文档中的配方:https://docs.python.org/2/library/decimal.html#recipes

from decimal import Decimal

def moneyfmt(value, places=3, curr='', sep=',', dp='.',
             pos='', neg='-', trailneg=''):
    [...]
numbers = [4.8, 49.723, 456.781, -72.18]
for x in numbers:
    value = moneyfmt(Decimal(x), places=2, pos=" ")
    print("{0:>10s}".format(value))

你会获得:

我是这样做的!

def alignDots(number):
    try:
        whole, dec = str(number).split('.')
        numWholeSpaces = 5 - len(whole) # Where 5 is number of spaces you want to theleft
        numDecSpaces   = 3 - len(dec)   # 3 is number of spaces to the right of the dot
        thousands = ' '* Math.abs(numWholeSpaces) + whole
        decimals  = dec + '0'*Math.abs(numDecSpaces)
        print thousands + '.' + decimals  
        return thousands + '.' + decimals  
    except:
        print "Failed to align dots of ",number
        return ' '*5+'ERROR'

我喜欢其他的解决方案，但需要一些具体的东西，想为什么不分享!

修复小数点

import decimal
numbers = [4.8, 49.723, 456.781, 50, -72.18, 12345.12345, 5000000000000]
dp = abs(min([decimal.Decimal(str(number)).as_tuple().exponent for number in numbers]))
width = max([len(str(int(number))) for number in numbers]) + dp + 1 #including .
for number in numbers:
    number = ("{:"+str(width)+"."+str(dp)+"f}").format(number)
    print number.rstrip('0').rstrip('.') if '.' in number else number

根据要求修正以考虑宽度:

numbers = [4.8, 49.723, 456.781, 50, -72.18]
width = max([len(str(number)) for number in numbers]) + 1
for number in numbers:
    number = ("{:"+str(width)+".4f}").format(number)
    print number.rstrip('0').rstrip('.') if '.' in number else number

编辑:如果你想包含整数

numbers = [4.8, 49.723, 456.781, 50, -72.18]
for number in numbers:
    number = "{:10.4f}".format(number)
    print number.rstrip('0').rstrip('.') if '.' in number else number

numbers = [4.8, 49.723, 456.781, -72.18]
for number in numbers:
    print "{:10.4f}".format(number).rstrip('0')

我知道得太晚了，但你也可以用数学，特别是对数的性质，来计算出你需要将所有数字填充到正确的小数点位置上的空格量。

from math import log10
nums = [4.8, 49.723, 456.781, -72.18, 5, 13]
def pre_spaces(nums):
    absmax = max([abs(max(nums)), abs(min(nums))])
    max_predot = int(log10(absmax))
    spaces = [' '*(max_predot-int(log10(abs(num))) - (1 if num<0 else 0)) for num in nums]
    return spaces
for s,n in zip(pre_spaces(nums), nums):
    print('{}{}'.format(s,n))

结果是:

您可以使用Python的[decimal][1]类型

有一个格式化货币值的配方。

为什么decimal类型:避免舍入问题，正确处理有效数字…

是的，有很多方法可以直接对齐浮点数和小数点。下面给出的两行代码是一个例子

"[20]:浮动= ((.022 12.645,544.5645 .54646,-554.56,-.2215,-546.5446])

In [21]: for xxx In floating: print "{0: 10.4f}".format(xxx)'

{0: 10.4f}中的0是每个浮点数项的维数。冒号后面的空格是可选的减号。10是小数点前的位数，4是小数点后的位数。这是我输出的JPG格式的结果

大多数答案首先生成一个字符串，然后在小数上进行分割以进行进一步的操作。还可以使用div和mod将数字分成整数部分和小数部分。解决方案可能是这样的:

for x in [4.8, 49.723, 456.781, -72.18,  5, 13]:
    print('[{:s}{}{:4s}]'.format('{:6.0f}'.format(x//1),
                                 '.' if x%1 else ' ',
                                 '{:0.4f}'.format(x%1).strip('.0')))

生产:

[     4.8   ]
[    49.723 ]
[   456.781 ]
[   -73.82  ]
[     5     ]
[    13     ]

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