将空数字类型转换为格式说明符

发布于 2025-01-05 07:30:42 字数 751 浏览 2 评论 0原文

有没有什么方法可以将“空”数字类型传递到数字格式说明符中，并让它打印与该说明符关联的空格？

IE。

n = "Scientific:"
v = 123.321
strfmt = "%5.4E"
output = "%s|"+strfmt+"|"
print output % (n, v)

>   Scientific:|1.2332E+02|

v = EMPTY
print output % (n, v)

>   Scientific:|          |

最终目标是处理不完整的行，而无需在循环时自适应更改格式字符串

perline = 3
n = ["1st:","2nd:","3rd:","4th:","5th:","6th:"]
v = [1.0, 2.0, 3.0, 4.0]

strfmt = "%5.4E"
output = ("%s|"+strfmt+"|  ")*perline+"\n"

for args in map(EMPTY,*[iter(n),iter(v)]*perline):
    print output % args

>   1st:|1.0000E+00|  2nd:|2.0000E+00|  3rd:|3.0000E+00|
>   4th:|4.0000E+00|  5th:|          |  6th:|          |

要执行上述代码，我将 EMPTY 替换为 None，尽管当您将 None 传递给格式字符串时会出现错误

原文

Is there any way to pass an "empty" numeric type into a numeric format specifier and have it print the blank space associated with that specifier?

ie.

n = "Scientific:"
v = 123.321
strfmt = "%5.4E"
output = "%s|"+strfmt+"|"
print output % (n, v)

>   Scientific:|1.2332E+02|

v = EMPTY
print output % (n, v)

>   Scientific:|          |

The ultimate goal is to deal with incomplete lines without adaptively changing the format string while looping

perline = 3
n = ["1st:","2nd:","3rd:","4th:","5th:","6th:"]
v = [1.0, 2.0, 3.0, 4.0]

strfmt = "%5.4E"
output = ("%s|"+strfmt+"|  ")*perline+"\n"

for args in map(EMPTY,*[iter(n),iter(v)]*perline):
    print output % args

>   1st:|1.0000E+00|  2nd:|2.0000E+00|  3rd:|3.0000E+00|
>   4th:|4.0000E+00|  5th:|          |  6th:|          |

To execute the above code I replaced EMPTY with None although that gives an error when you pass None to the format string

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眼前雾蒙蒙 2025-01-12 07:30:42

您将无法仅通过数字格式来完成此操作。你需要一些逻辑：

strfmt = "%5.4E"
v = None

if v in [None, '']:
  output = "%s|          |"
else:
  output = "%s|"+strfmt+"|"

You won't be able to do it with just number formatting. You'll need some logic:

strfmt = "%5.4E"
v = None

if v in [None, '']:
  output = "%s|          |"
else:
  output = "%s|"+strfmt+"|"

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暮光沉寂 2025-01-12 07:30:42

简短回答：对于 n 个位置宽的字段，'% n s' % '' 就可以了。也就是说，'%15s' % '' 将打印 15 个空格。

长答案：当您在格式说明符中指定宽度时，第一个数字是最小宽度整个字段。不过，"%5.4E" 将占据超过 5 个位置，因为仅指数就已经占据了 3 个位置。

指定类似 "%12.4E" 的内容并获得一致的结果。对于缺失值，请指定 "%12s" 并传递空字符串（而不是 None）而不是值。

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离笑几人歌 2025-01-12 07:30:42

一种选择是将 output 中的数字格式替换为 %s，然后在替换元组中创建正确的数字输出或空白空间，具体取决于 >v 是：

>>> v = 123.321
>>> "%s|%s|" % (n, "%5.4E" % v if v is not None else " "*10)
'Scientific:|1.2332E+02|'
>>> v = None
>>> "%s|%s|" % (n, "%5.4E" % v if v is not None else " "*10)
'Scientific:|          |'

您可以通过一个简单的闭包使其功能更强大：

def make_default_blank_formatter(format, default):
    def default_blank_formatter(v):
        return format % v if v is not None else default
    return default_blank_formatter

scientific_formatter = make_default_blank_formatter("%5.4E", " "*10)

>>> v = 123.321
>>> "%s|%s|" % (n, scientific_formatter(v))
'Scientific:|1.2332E+02|'
>>> v = None
>>> "%s|%s|" % (n, scientific_formatter(v))
'Scientific:|          |'

编辑：以下是您如何重新编写打印代码以使用它：

perline = 3
n = ["1st:","2nd:","3rd:","4th:","5th:","6th:"]
v = [1.0, 2.0, 3.0, 4.0, None, None]

strfmt = "%5.4E"
output = ("%s|%s|  ")*perline

for args in zip(*[iter(n),iter(map(scientific_formatter, v))]*perline):
    print output % args

One option would be to replace the numeric format in output with %s, and then in the replacing tuple create either the correct numeric output or the empty space depending on what v is:

>>> v = 123.321
>>> "%s|%s|" % (n, "%5.4E" % v if v is not None else " "*10)
'Scientific:|1.2332E+02|'
>>> v = None
>>> "%s|%s|" % (n, "%5.4E" % v if v is not None else " "*10)
'Scientific:|          |'

You could make this a little more functional with a simple closure:

def make_default_blank_formatter(format, default):
    def default_blank_formatter(v):
        return format % v if v is not None else default
    return default_blank_formatter

scientific_formatter = make_default_blank_formatter("%5.4E", " "*10)

>>> v = 123.321
>>> "%s|%s|" % (n, scientific_formatter(v))
'Scientific:|1.2332E+02|'
>>> v = None
>>> "%s|%s|" % (n, scientific_formatter(v))
'Scientific:|          |'

edit: Here is how you could rework your printing code to use this:

perline = 3
n = ["1st:","2nd:","3rd:","4th:","5th:","6th:"]
v = [1.0, 2.0, 3.0, 4.0, None, None]

strfmt = "%5.4E"
output = ("%s|%s|  ")*perline

for args in zip(*[iter(n),iter(map(scientific_formatter, v))]*perline):
    print output % args

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情场扛把子 2025-01-12 07:30:42

您可以使用 itertools.izip_longest 将 n 和 v 中的项目配对，并使用空字符串填充缺失值。

要获得 v 中的项目的正确格式，您可以使用 strfmt 预先格式化浮点数，并使用类似

output = "%s|%10s|  "

连接 n 中的项目的方法代码>和<代码>v。通过这种方式，v 中的项目现在都是字符串，包括空字符串，而不是浮点数和空字符串的混合。

import itertools

perline = 3
n = ["1st:", "2nd:", "3rd:", "4th:", "5th:", "6th:"]
v = [1.0, 2.0, 3.0, 4.0]

strfmt = "%5.4E"
v = (strfmt % val for val in v)
output = "%s|%10s|  "

pairs = itertools.izip_longest(n, v, fillvalue = '')
items = (output % (place, val) for place, val in pairs)
for row in zip(*[items]*perline):
    print(''.join(row))

产量

1st:|1.0000E+00|  2nd:|2.0000E+00|  3rd:|3.0000E+00|  
4th:|4.0000E+00|  5th:|          |  6th:|          |

You could use itertools.izip_longest to pair the items in n and v, and fill in missing values with the empty string.

To get the proper formatting for the items in v, you could pre-format the floats with strfmt, and use something like

output = "%s|%10s|  "

to join the items in n and v. By doing it this way, the items in v are now all strings, including the empty strings, and not a mixture of floats and empty strings.

import itertools

perline = 3
n = ["1st:", "2nd:", "3rd:", "4th:", "5th:", "6th:"]
v = [1.0, 2.0, 3.0, 4.0]

strfmt = "%5.4E"
v = (strfmt % val for val in v)
output = "%s|%10s|  "

pairs = itertools.izip_longest(n, v, fillvalue = '')
items = (output % (place, val) for place, val in pairs)
for row in zip(*[items]*perline):
    print(''.join(row))

yields

1st:|1.0000E+00|  2nd:|2.0000E+00|  3rd:|3.0000E+00|  
4th:|4.0000E+00|  5th:|          |  6th:|          |

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余罪 2025-01-12 07:30:42

如上所述，只需在打印之前将数字转换为字符串，然后用相同长度的空字符串替换缺失的数字。这实际上可以写得非常紧凑，但是我永远不会在严肃的代码中使用它：

from itertools import izip_longest, groupby, chain

perline = 3
n = ["1st:","2nd:","3rd:","4th:","5th:","6th:"]
v = [1.0, 2.0, 3.0, 4.0]

strfmt = "%5.4E"
output = ("%s|%s|  ")*perline

for _, args in groupby(enumerate(chain(*izip_longest(n, (strfmt % x for x in v),
                                                     fillvalue = ' '*len(strfmt % 0)))),
                       lambda (i, v): i/perline/2):
    print output % zip(*args)[1]

最后一个循环的更具可读性和可靠性的版本：

l = ()
for pair in izip_longest(n, (strfmt % x for x in v),
                         fillvalue = ' '*len(strfmt % 0)):
    l += pair
    if len(l) == perline*2:
        print output % l
        l = ()

As suggested above, just convert your numbers to strings before printing, and replace missing numbers with empty strings of the equal length. This can be actually written very compact, however I would never use this in serious code:

from itertools import izip_longest, groupby, chain

perline = 3
n = ["1st:","2nd:","3rd:","4th:","5th:","6th:"]
v = [1.0, 2.0, 3.0, 4.0]

strfmt = "%5.4E"
output = ("%s|%s|  ")*perline

for _, args in groupby(enumerate(chain(*izip_longest(n, (strfmt % x for x in v),
                                                     fillvalue = ' '*len(strfmt % 0)))),
                       lambda (i, v): i/perline/2):
    print output % zip(*args)[1]

More readable and reliable version of the last loop:

l = ()
for pair in izip_longest(n, (strfmt % x for x in v),
                         fillvalue = ' '*len(strfmt % 0)):
    l += pair
    if len(l) == perline*2:
        print output % l
        l = ()

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