部分字符串格式

Question

是否可以使用高级字符串格式化方法进行部分字符串格式化，类似于字符串模板safe_substitute()函数？

例如：

s = '{foo} {bar}'
s.format(foo='FOO') #Problem: raises KeyError 'bar'

Answer 1

如果您知道格式化的是什么顺序：

s = '{foo} {{bar}}'

像这样使用：

ss = s.format(foo='FOO') 
print ss 
>>> 'FOO {bar}'

print ss.format(bar='BAR')
>>> 'FOO BAR'

您无法同时指定foo和bar - 您必须按顺序执行此操作。

Answer 2

您可以使用partial中的functools函数，该函数简短，最具可读性，并且最能说明编码人员的意图：

from functools import partial

s = partial("{foo} {bar}".format, foo="FOO")
print s(bar="BAR")
# FOO BAR

Answer 3

您可以通过覆盖映射将其欺骗为部分格式化：

import string

class FormatDict(dict):
    def __missing__(self, key):
        return "{" + key + "}"

s = '{foo} {bar}'
formatter = string.Formatter()
mapping = FormatDict(foo='FOO')
print(formatter.vformat(s, (), mapping))

印刷

FOO {bar}

当然，这个基本实现仅适用于基本情况。

Answer 4

.format()的这种限制 - 无法进行部分替换 - 一直困扰着我。

在评估编写自定义Formatter类之后，如此处的许多答案所述，甚至考虑使用第三方软件包，例如lazy_format，我发现了一个更简单的内置解决方案：Template strings < / p>

它提供了类似的功能，但也通过safe_substitute()方法提供部分替换。模板字符串需要有$前缀（感觉有点奇怪 - 但我觉得整体解决方案更好）。

import string
template = string.Template('${x} ${y}')
try:
  template.substitute({'x':1}) # raises KeyError
except KeyError:
  pass

# but the following raises no error
partial_str = template.safe_substitute({'x':1}) # no error

# partial_str now contains a string with partial substitution
partial_template = string.Template(partial_str)
substituted_str = partial_template.safe_substitute({'y':2}) # no error
print substituted_str # prints '12'

基于此形成了一个便利包装：

class StringTemplate(object):
    def __init__(self, template):
        self.template = string.Template(template)
        self.partial_substituted_str = None

    def __repr__(self):
        return self.template.safe_substitute()

    def format(self, *args, **kws):
        self.partial_substituted_str = self.template.safe_substitute(*args, **kws)
        self.template = string.Template(self.partial_substituted_str)
        return self.__repr__()


>>> s = StringTemplate('${x}${y}')
>>> s
'${x}${y}'
>>> s.format(x=1)
'1${y}'
>>> s.format({'y':2})
'12'
>>> print s
12

同样是基于Sven答案的包装器，它使用默认的字符串格式：

class StringTemplate(object):
    class FormatDict(dict):
        def __missing__(self, key):
            return "{" + key + "}"

    def __init__(self, template):
        self.substituted_str = template
        self.formatter = string.Formatter()

    def __repr__(self):
        return self.substituted_str

    def format(self, *args, **kwargs):
        mapping = StringTemplate.FormatDict(*args, **kwargs)
        self.substituted_str = self.formatter.vformat(self.substituted_str, (), mapping)

Answer 5

不确定这是否可以作为快速解决方法，但

怎么样

s = '{foo} {bar}'
s.format(foo='FOO', bar='{bar}')

？ ：）

Answer 6

如果您定义自己的Formatter覆盖get_value方法，则可以使用该方法将未定义的字段名称映射到您想要的任何内容：

http://docs.python.org/library/string.html#string.Formatter.get_value

例如，如果bar不在kwargs中，您可以将"{bar}"映射到bar。

但是，这需要使用Formatter对象的format()方法，而不是字符串的format()方法。

Answer 7

>>> 'fd:{uid}:{{topic_id}}'.format(uid=123)
'fd:123:{topic_id}'

试试这个。

Answer 8

感谢Amber的评论，我提出了这个问题：

import string

try:
    # Python 3
    from _string import formatter_field_name_split
except ImportError:
    formatter_field_name_split = str._formatter_field_name_split


class PartialFormatter(string.Formatter):
    def get_field(self, field_name, args, kwargs):
        try:
            val = super(PartialFormatter, self).get_field(field_name, args, kwargs)
        except (IndexError, KeyError, AttributeError):
            first, _ = formatter_field_name_split(field_name)
            val = '{' + field_name + '}', first
        return val

Answer 9

对我而言，这已经足够了：

>>> ss = 'dfassf {} dfasfae efaef {} fds'
>>> nn = ss.format('f1', '{}')
>>> nn
'dfassf f1 dfasfae efaef {} fds'
>>> n2 = nn.format('whoa')
>>> n2
'dfassf f1 dfasfae efaef whoa fds'

Answer 10

我发现的所有解决方案似乎在使用更高级的规格或转换选项方面均存在问题。 @SvenMarnach的FormatPlaceholder非常聪明，但是不能强制使用（例如{a!s:>2s}），因为它调用__str__方法（在此示例中）而不是__format__，并且您将丢失任何其他格式。

这是我最终得到的结果，并且其中包括一些关键功能：

sformat('The {} is {}', 'answer')
'The answer is {}'

sformat('The answer to {question!r} is {answer:0.2f}', answer=42)
'The answer to {question!r} is 42.00'

sformat('The {} to {} is {:0.{p}f}', 'answer', 'everything', p=4)
'The answer to everything is {:0.4f}'

• 提供与str.format类似的界面（不仅仅是映射）
• 支持更复杂的格式选项：
  • 胁迫{k!s} {!r}
  • 嵌套{k:>{size}}
  • getattr {k.foo}
  • getitem {k[0]}
  • 强制+格式化{k!s:>{size}}

import string


class SparseFormatter(string.Formatter):
    """
    A modified string formatter that handles a sparse set of format
    args/kwargs.
    """

    # re-implemented this method for python2/3 compatibility
    def vformat(self, format_string, args, kwargs):
        used_args = set()
        result, _ = self._vformat(format_string, args, kwargs, used_args, 2)
        self.check_unused_args(used_args, args, kwargs)
        return result

    def _vformat(self, format_string, args, kwargs, used_args, recursion_depth,
                 auto_arg_index=0):
        if recursion_depth < 0:
            raise ValueError('Max string recursion exceeded')
        result = []
        for literal_text, field_name, format_spec, conversion in \
                self.parse(format_string):

            orig_field_name = field_name

            # output the literal text
            if literal_text:
                result.append(literal_text)

            # if there's a field, output it
            if field_name is not None:
                # this is some markup, find the object and do
                #  the formatting

                # handle arg indexing when empty field_names are given.
                if field_name == '':
                    if auto_arg_index is False:
                        raise ValueError('cannot switch from manual field '
                                         'specification to automatic field '
                                         'numbering')
                    field_name = str(auto_arg_index)
                    auto_arg_index += 1
                elif field_name.isdigit():
                    if auto_arg_index:
                        raise ValueError('cannot switch from manual field '
                                         'specification to automatic field '
                                         'numbering')
                    # disable auto arg incrementing, if it gets
                    # used later on, then an exception will be raised
                    auto_arg_index = False

                # given the field_name, find the object it references
                #  and the argument it came from
                try:
                    obj, arg_used = self.get_field(field_name, args, kwargs)
                except (IndexError, KeyError):
                    # catch issues with both arg indexing and kwarg key errors
                    obj = orig_field_name
                    if conversion:
                        obj += '!{}'.format(conversion)
                    if format_spec:
                        format_spec, auto_arg_index = self._vformat(
                            format_spec, args, kwargs, used_args,
                            recursion_depth, auto_arg_index=auto_arg_index)
                        obj += ':{}'.format(format_spec)
                    result.append('{' + obj + '}')
                else:
                    used_args.add(arg_used)

                    # do any conversion on the resulting object
                    obj = self.convert_field(obj, conversion)

                    # expand the format spec, if needed
                    format_spec, auto_arg_index = self._vformat(
                        format_spec, args, kwargs,
                        used_args, recursion_depth-1,
                        auto_arg_index=auto_arg_index)

                    # format the object and append to the result
                    result.append(self.format_field(obj, format_spec))

        return ''.join(result), auto_arg_index


def sformat(s, *args, **kwargs):
    # type: (str, *Any, **Any) -> str
    """
    Sparse format a string.

    Parameters
    ----------
    s : str
    args : *Any
    kwargs : **Any

    Examples
    --------
    >>> sformat('The {} is {}', 'answer')
    'The answer is {}'

    >>> sformat('The answer to {question!r} is {answer:0.2f}', answer=42)
    'The answer to {question!r} is 42.00'

    >>> sformat('The {} to {} is {:0.{p}f}', 'answer', 'everything', p=4)
    'The answer to everything is {:0.4f}'

    Returns
    -------
    str
    """
    return SparseFormatter().format(s, *args, **kwargs)

在编写了一些关于我希望这种方法如何运行的测试之后，我发现了各种实现的问题。如果有人发现他们有见地，它们就会在下面。

import pytest


def test_auto_indexing():
    # test basic arg auto-indexing
    assert sformat('{}{}', 4, 2) == '42'
    assert sformat('{}{} {}', 4, 2) == '42 {}'


def test_manual_indexing():
    # test basic arg indexing
    assert sformat('{0}{1} is not {1} or {0}', 4, 2) == '42 is not 2 or 4'
    assert sformat('{0}{1} is {3} {1} or {0}', 4, 2) == '42 is {3} 2 or 4'


def test_mixing_manualauto_fails():
    # test mixing manual and auto args raises
    with pytest.raises(ValueError):
        assert sformat('{!r} is {0}{1}', 4, 2)


def test_kwargs():
    # test basic kwarg
    assert sformat('{base}{n}', base=4, n=2) == '42'
    assert sformat('{base}{n}', base=4, n=2, extra='foo') == '42'
    assert sformat('{base}{n} {key}', base=4, n=2) == '42 {key}'


def test_args_and_kwargs():
    # test mixing args/kwargs with leftovers
    assert sformat('{}{k} {v}', 4, k=2) == '42 {v}'

    # test mixing with leftovers
    r = sformat('{}{} is the {k} to {!r}', 4, 2, k='answer')
    assert r == '42 is the answer to {!r}'


def test_coercion():
    # test coercion is preserved for skipped elements
    assert sformat('{!r} {k!r}', '42') == "'42' {k!r}"


def test_nesting():
    # test nesting works with or with out parent keys
    assert sformat('{k:>{size}}', k=42, size=3) == ' 42'
    assert sformat('{k:>{size}}', size=3) == '{k:>3}'


@pytest.mark.parametrize(
    ('s', 'expected'),
    [
        ('{a} {b}', '1 2.0'),
        ('{z} {y}', '{z} {y}'),
        ('{a} {a:2d} {a:04d} {y:2d} {z:04d}', '1  1 0001 {y:2d} {z:04d}'),
        ('{a!s} {z!s} {d!r}', '1 {z!s} {\'k\': \'v\'}'),
        ('{a!s:>2s} {z!s:>2s}', ' 1 {z!s:>2s}'),
        ('{a!s:>{a}s} {z!s:>{z}s}', '1 {z!s:>{z}s}'),
        ('{a.imag} {z.y}', '0 {z.y}'),
        ('{e[0]:03d} {z[0]:03d}', '042 {z[0]:03d}'),
    ],
    ids=[
        'normal',
        'none',
        'formatting',
        'coercion',
        'formatting+coercion',
        'nesting',
        'getattr',
        'getitem',
    ]
)
def test_sformat(s, expected):
    # test a bunch of random stuff
    data = dict(
        a=1,
        b=2.0,
        c='3',
        d={'k': 'v'},
        e=[42],
    )
    assert expected == sformat(s, **data)

Answer 11

这是一个基于正则表达式的轻度漏洞解决方案。请注意，这将不与{foo:{width}}之类的嵌套格式说明符一起使用，但是它确实解决了其他答案存在的一些问题。

def partial_format(s, **kwargs):
    parts = re.split(r'(\{[^}]*\})', s)
    for k, v in kwargs.items():
        for idx, part in enumerate(parts):
            if re.match(rf'\{{{k}[!:}}]', part):  # Placeholder keys must always be followed by '!', ':', or the closing '}'
                parts[idx] = parts[idx].format_map({k: v})
    return ''.join(parts)

# >>> partial_format('{foo} {bar:1.3f}', foo='FOO')
# 'FOO {bar:1.3f}'
# >>> partial_format('{foo} {bar:1.3f}', bar=1)
# '{foo} 1.000'

Answer 12

我的建议如下（使用Python3.6测试）：

class Lazymap(object):
       def __init__(self, **kwargs):
           self.dict = kwargs

       def __getitem__(self, key):
           return self.dict.get(key, "".join(["{", key, "}"]))


s = '{foo} {bar}'

s.format_map(Lazymap(bar="FOO"))
# >>> '{foo} FOO'

s.format_map(Lazymap(bar="BAR"))
# >>> '{foo} BAR'

s.format_map(Lazymap(bar="BAR", foo="FOO", baz="BAZ"))
# >>> 'FOO BAR'

更新： 此处显示了一种更优雅的方法（子类dict和重载__missing__(self, key)）：https://stackoverflow.com/a/17215533/333403

Answer 13

TL;DR：问题：如果 defaultdict 未设置，{foobar[a]} 将失败 foobar：

from collections import defaultdict

text = "{bar}, {foo}, {foobar[a]}" # {bar} is set, {foo} is "", {foobar[a]} fails
text.format_map(defaultdict(str, bar="A")) # TypeError: string indices must be integers

解决方案：从编辑复制DefaultWrapper类，然后：

text = "{bar}, {foo}, {foobar[a]}"
text.format_map(DefaultWrapper(bar="A")) # "A, , " (missing replaced with empty str)

# Even this works:
foobar = {"c": "C"}
text = "{foobar[a]}, {foobar[c]}"
text.format_map(DefaultWrapper(foobar=foobar)) # ", C" missing indices are also replaced

---

请注意，索引和属性访问在已发布的解决方案之一中不起作用。以下代码引发 TypeError: string indices must be integers.

from collections import defaultdict

text = "{foo} '{bar[index]}'"
text.format_map(defaultdict(str, foo="FOO")) # raises a TypeError

要解决此问题，可以将 collections.defaultdict 解决方案与支持索引的自定义默认值对象一起使用。 DefaultWrapper 对象在索引和属性访问时返回自身，允许无限次索引/使用属性而不会出错。

请注意，这可以扩展以允许包含部分请求值的容器。查看下面的编辑。

class DefaultWrapper:
    def __repr__(self):
        return "Empty default value"
    
    def __str__(self):
        return ""
    
    def __format__(self, format_spec):
        return ""
    
    def __getattr__(self, name):
        return self
    
    def __getitem__(self, name):
        return self
    
    def __contains__(self, name):
        return True

text = "'{foo}', '{bar[index][i]}'"
print(text.format_map(defaultdict(DefaultWrapper, foo="FOO")))
# 'FOO', ''

---

编辑：部分填充的容器

上述类可以扩展以支持部分填充的容器。所以例如

text = "'{foo[a]}', '{foo[b]}'"
foo = {"a": "A"}

print(text.format_map(defaultdict(DefaultWrapper, foo=foo)))
# KeyError: 'b'

这个想法是用 defaultdict 完全替换 DefaultWrapper。 DefaultWrapper 对象环绕 container 返回容器请求的值（用 DefaultWrapper 对象包装）或容器作为字符串。这种方式模仿了地图的无限深度，但返回了所有当前值。

添加 kwargs 只是为了方便。这样看起来更像是 defaultdict 解决方案。

class DefaultWrapper:
    """A wrapper around the `container` to allow accessing with a default value."""
    ignore_str_format_errors = True

    def __init__(self, container="", **kwargs):
        self.container = container
        self.kwargs = kwargs

    def __repr__(self):
        return "DefaultWrapper around '{}'".format(repr(self.container))

    def __str__(self):
        return str(self.container)
    
    def __format__(self, format_spec):
        try:
            return self.container.__format__(format_spec)
        except TypeError as e:
            if DefaultWrapper.ignore_str_format_errors or self.container == "":
                return str(self)
            else:
                raise e

    def __getattr__(self, name):
        try:
            return DefaultWrapper(getattr(self.container, name))
        except AttributeError:
            return DefaultWrapper()

    def __getitem__(self, name):
        try:
            return DefaultWrapper(self.container[name])
        except (TypeError, LookupError):
            try:
                return DefaultWrapper(self.kwargs[name])
            except (TypeError, LookupError):
                return DefaultWrapper()
        
    def __contains__(self, name):
        return True

现在所有显示的示例都没有错误：

text = "'{foo[a]}', '{foo[b]}'"
foo = {"a": "A"}
print(text.format_map(DefaultWrapper(foo=foo)))
# 'A', ''

text = "'{foo}', '{bar[index][i]}', '{foobar[a]}', '{foobar[b]}'"
print(text.format_map(DefaultWrapper(foo="Foo", foobar={"a": "A"})))
# 'FOO', '', 'A', ''

# the old way still works the same as before
from collections import defaultdict
text = "'{foo}', '{bar[index][i]}'"
print(text.format_map(defaultdict(DefaultWrapper, foo="FOO")))
# 'FOO', ''

Answer 14

阅读@Sam Bourne评论，我修改了@SvenMarnach的code 可以在不编写自定义解析器的情况下正确使用强制（例如{a!s:>2s}）。 基本思想不是转换为字符串，而是用强制标签将丢失的键连接起来。

import string
class MissingKey(object):
    def __init__(self, key):
        self.key = key

    def __str__(self):  # Supports {key!s}
        return MissingKeyStr("".join([self.key, "!s"]))

    def __repr__(self):  # Supports {key!r}
        return MissingKeyStr("".join([self.key, "!r"]))

    def __format__(self, spec): # Supports {key:spec}
        if spec:
            return "".join(["{", self.key, ":", spec, "}"])
        return "".join(["{", self.key, "}"])

    def __getitem__(self, i): # Supports {key[i]}
        return MissingKey("".join([self.key, "[", str(i), "]"]))

    def __getattr__(self, name): # Supports {key.name}
        return MissingKey("".join([self.key, ".", name]))


class MissingKeyStr(MissingKey, str):
    def __init__(self, key):
        if isinstance(key, MissingKey):
            self.key = "".join([key.key, "!s"])
        else:
            self.key = key

class SafeFormatter(string.Formatter):
    def __init__(self, default=lambda k: MissingKey(k)):
        self.default=default

    def get_value(self, key, args, kwds):
        if isinstance(key, str):
            return kwds.get(key, self.default(key))
        else:
            return super().get_value(key, args, kwds)

像这样使用（例如）

SafeFormatter().format("{a:<5} {b:<10}", a=10)

以下测试（受@ norok2的启发）在两种情况下根据上述类检查传统format_map和safe_format_map的输出：提供正确的关键字或不提供关键字。

def safe_format_map(text, source):
    return SafeFormatter().format(text, **source)

test_texts = (
    '{a} ',             # simple nothing useful in source
    '{a:5d}',       # formatting
    '{a!s}',        # coercion
    '{a!s:>{a}s}',  # formatting and coercion
    '{a:0{a}d}',    # nesting
    '{d[x]}',       # indexing
    '{d.values}',   # member
)

source = dict(a=10,d=dict(x='FOO'))
funcs = [safe_format_map,
         str.format_map
         #safe_format_alt  # Version based on parsing (See @norok2)
         ]
n = 18
for text in test_texts:
    # full_source = {**dict(b='---', f=dict(g='Oh yes!')), **source}
    # print('{:>{n}s} :   OK   : '.format('str.format_map', n=n) + text.format_map(full_source))
    print("Testing:", text)
    for func in funcs:
        try:
            print(f'{func.__name__:>{n}s} : OK\t\t\t: ' + func(text, dict()))
        except:
            print(f'{func.__name__:>{n}s} : FAILED')

        try:
            print(f'{func.__name__:>{n}s} : OK\t\t\t: ' + func(text, source))
        except:
            print(f'{func.__name__:>{n}s} : FAILED')

哪个输出

Testing: {a} 
   safe_format_map : OK         : {a} 
   safe_format_map : OK         : 10 
        format_map : FAILED
        format_map : OK         : 10 
Testing: {a:5d}
   safe_format_map : OK         : {a:5d}
   safe_format_map : OK         :    10
        format_map : FAILED
        format_map : OK         :    10
Testing: {a!s}
   safe_format_map : OK         : {a!s}
   safe_format_map : OK         : 10
        format_map : FAILED
        format_map : OK         : 10
Testing: {a!s:>{a}s}
   safe_format_map : OK         : {a!s:>{a}s}
   safe_format_map : OK         :         10
        format_map : FAILED
        format_map : OK         :         10
Testing: {a:0{a}d}
   safe_format_map : OK         : {a:0{a}d}
   safe_format_map : OK         : 0000000010
        format_map : FAILED
        format_map : OK         : 0000000010
Testing: {d[x]}
   safe_format_map : OK         : {d[x]}
   safe_format_map : OK         : FOO
        format_map : FAILED
        format_map : OK         : FOO
Testing: {d.values}
   safe_format_map : OK         : {d.values}
   safe_format_map : OK         : <built-in method values of dict object at 0x7fe61e230af8>
        format_map : FAILED
        format_map : OK         : <built-in method values of dict object at 0x7fe61e230af8>

Answer 15

如果您进行了大量的模板处理，并且发现Python的内置字符串模板功能不足或笨拙，请查看Jinja2。

从文档中

Jinja是一种现代且设计友好的Python模板语言，以Django的模板为模型。

Answer 16

我喜欢@ sven-marnach的答案。我的答案只是它的扩展版本。它允许非关键字格式，并忽略多余的键。以下是用法示例（函数名称是对python 3.6 f字符串格式的引用）：

# partial string substitution by keyword
>>> f('{foo} {bar}', foo="FOO")
'FOO {bar}'

# partial string substitution by argument
>>> f('{} {bar}', 1)
'1 {bar}'

>>> f('{foo} {}', 1)
'{foo} 1'

# partial string substitution with arguments and keyword mixed
>>> f('{foo} {} {bar} {}', '|', bar='BAR')
'{foo} | BAR {}'

# partial string substitution with extra keyword
>>> f('{foo} {bar}', foo="FOO", bro="BRO")
'FOO {bar}'

# you can simply 'pour out' your dictionary to format function
>>> kwargs = {'foo': 'FOO', 'bro': 'BRO'}
>>> f('{foo} {bar}', **kwargs)
'FOO {bar}'

---

这是我的代码：

from string import Formatter


class FormatTuple(tuple):
    def __getitem__(self, key):
        if key + 1 > len(self):
            return "{}"
        return tuple.__getitem__(self, key)


class FormatDict(dict):
    def __missing__(self, key):
        return "{" + key + "}"


def f(string, *args, **kwargs):
    """
    String safe substitute format method.
    If you pass extra keys they will be ignored.
    If you pass incomplete substitute map, missing keys will be left unchanged.
    :param string:
    :param kwargs:
    :return:

    >>> f('{foo} {bar}', foo="FOO")
    'FOO {bar}'
    >>> f('{} {bar}', 1)
    '1 {bar}'
    >>> f('{foo} {}', 1)
    '{foo} 1'
    >>> f('{foo} {} {bar} {}', '|', bar='BAR')
    '{foo} | BAR {}'
    >>> f('{foo} {bar}', foo="FOO", bro="BRO")
    'FOO {bar}'
    """
    formatter = Formatter()
    args_mapping = FormatTuple(args)
    mapping = FormatDict(kwargs)
    return formatter.vformat(string, args_mapping, mapping)

Answer 17

如果您想解开字典以将参数传递给format，as in this related question，则可以使用以下方法。

首先假定字符串s与此问题相同：

s = '{foo} {bar}'

，值由以下字典给出：

replacements = {'foo': 'FOO'}

显然这行不通：

s.format(**replacements)
#---------------------------------------------------------------------------
#KeyError                                  Traceback (most recent call last)
#<ipython-input-29-ef5e51de79bf> in <module>()
#----> 1 s.format(**replacements)
#
#KeyError: 'bar'

但是，您可以先get a set of all of the named arguments from s并创建一个字典，该字典将参数映射到用大括号括起来的自身：

from string import Formatter
args = {x[1]:'{'+x[1]+'}' for x in Formatter().parse(s)}
print(args)
#{'foo': '{foo}', 'bar': '{bar}'}

现在使用args字典填写replacements中缺少的键。对于python 3.5+，您可以do this in a single expression：

new_s = s.format(**{**args, **replacements}}
print(new_s)
#'FOO {bar}'

对于旧版本的python，您可以调用update：

args.update(replacements)
print(s.format(**args))
#'FOO {bar}'

Answer 18

在测试了here和there的最有希望的解决方案之后，我意识到它们没有真正满足以下要求：

1. 严格遵守str.format_map()识别的模板语法；
2. 能够保留复杂的格式，即完全支持Format Mini-Language

因此，我编写了自己的解决方案，它满足了上述要求。 （ EDIT ：现在，@ SvenMarnach的版本（如该答案中所述）似乎可以解决我所需要的一些极端情况。）

基本上，我最终解析了模板字符串，找到了匹配的嵌套{.*?}组（使用find_all()帮助函数），并使用{ {1}}，同时捕捉到任何潜在的str.format_map()。

KeyError

def find_all(
        text,
        pattern,
        overlap=False):
    """
    Find all occurrencies of the pattern in the text.

    Args:
        text (str|bytes|bytearray): The input text.
        pattern (str|bytes|bytearray): The pattern to find.
        overlap (bool): Detect overlapping patterns.

    Yields:
        position (int): The position of the next finding.
    """
    len_text = len(text)
    offset = 1 if overlap else (len(pattern) or 1)
    i = 0
    while i < len_text:
        i = text.find(pattern, i)
        if i >= 0:
            yield i
            i += offset
        else:
            break

def matching_delimiters(
        text,
        l_delim,
        r_delim,
        including=True):
    """
    Find matching delimiters in a sequence.

    The delimiters are matched according to nesting level.

    Args:
        text (str|bytes|bytearray): The input text.
        l_delim (str|bytes|bytearray): The left delimiter.
        r_delim (str|bytes|bytearray): The right delimiter.
        including (bool): Include delimeters.

    yields:
        result (tuple[int]): The matching delimiters.
    """
    l_offset = len(l_delim) if including else 0
    r_offset = len(r_delim) if including else 0
    stack = []

    l_tokens = set(find_all(text, l_delim))
    r_tokens = set(find_all(text, r_delim))
    positions = l_tokens.union(r_tokens)
    for pos in sorted(positions):
        if pos in l_tokens:
            stack.append(pos + 1)
        elif pos in r_tokens:
            if len(stack) > 0:
                prev = stack.pop()
                yield (prev - l_offset, pos + r_offset, len(stack))
            else:
                raise ValueError(
                    'Found `{}` unmatched right token(s) `{}` (position: {}).'
                        .format(len(r_tokens) - len(l_tokens), r_delim, pos))
    if len(stack) > 0:
        raise ValueError(
            'Found `{}` unmatched left token(s) `{}` (position: {}).'
                .format(
                len(l_tokens) - len(r_tokens), l_delim, stack.pop() - 1))

（FlyingCircus中也提供此代码-免责声明：我是它的主要作者。）

---

此代码的用法为：

def safe_format_map(
        text,
        source):
    """
    Perform safe string formatting from a mapping source.

    If a value is missing from source, this is simply ignored, and no
    `KeyError` is raised.

    Args:
        text (str): Text to format.
        source (Mapping|None): The mapping to use as source.
            If None, uses caller's `vars()`.

    Returns:
        result (str): The formatted text.
    """
    stack = []
    for i, j, depth in matching_delimiters(text, '{', '}'):
        if depth == 0:
            try:
                replacing = text[i:j].format_map(source)
            except KeyError:
                pass
            else:
                stack.append((i, j, replacing))
    result = ''
    i, j = len(text), 0
    while len(stack) > 0:
        last_i = i
        i, j, replacing = stack.pop()
        result = replacing + text[j:last_i] + result
    if i > 0:
        result = text[0:i] + result
    return result

---

让我们将其与我最喜欢的解决方案（由@SvenMarnach共享他的代码here和there进行比较）

print(safe_format_map('{a} {b} {c}', dict(a=-A-)))
# -A- {b} {c}

以下是一些测试：

import string


class FormatPlaceholder:
    def __init__(self, key):
        self.key = key
    def __format__(self, spec):
        result = self.key
        if spec:
            result += ":" + spec
        return "{" + result + "}"
    def __getitem__(self, index):
        self.key = "{}[{}]".format(self.key, index)
        return self
    def __getattr__(self, attr):
        self.key = "{}.{}".format(self.key, attr)
        return self


class FormatDict(dict):
    def __missing__(self, key):
        return FormatPlaceholder(key)


def safe_format_alt(text, source):
    formatter = string.Formatter()
    return formatter.vformat(text, (), FormatDict(source))

以及使其运行的代码：

test_texts = (
    '{b} {f}',  # simple nothing useful in source
    '{a} {b}',  # simple
    '{a} {b} {c:5d}',  # formatting
    '{a} {b} {c!s}',  # coercion
    '{a} {b} {c!s:>{a}s}',  # formatting and coercion
    '{a} {b} {c:0{a}d}',  # nesting
    '{a} {b} {d[x]}',  # dicts (existing in source)
    '{a} {b} {e.index}',  # class (existing in source)
    '{a} {b} {f[g]}',  # dict (not existing in source)
    '{a} {b} {f.values}',  # class (not existing in source)

)
source = dict(a=4, c=101, d=dict(x='FOO'), e=[])

导致：

funcs = safe_format_map, safe_format_alt

n = 18
for text in test_texts:
    full_source = {**dict(b='---', f=dict(g='Oh yes!')), **source}
    print('{:>{n}s} :   OK   : '.format('str.format_map', n=n) + text.format_map(full_source))
    for func in funcs:
        try:
            print(f'{func.__name__:>{n}s} :   OK   : ' + func(text, source))
        except:
            print(f'{func.__name__:>{n}s} : FAILED : {text}')

如您所见，更新的版本现在似乎可以很好地处理早期版本曾经失败的特殊情况。

---

时间上，它们大约在彼此的50％，取决于要格式化的实际 str.format_map : OK : --- {'g': 'Oh yes!'} safe_format_map : OK : {b} {f} safe_format_alt : OK : {b} {f} str.format_map : OK : 4 --- safe_format_map : OK : 4 {b} safe_format_alt : OK : 4 {b} str.format_map : OK : 4 --- 101 safe_format_map : OK : 4 {b} 101 safe_format_alt : OK : 4 {b} 101 str.format_map : OK : 4 --- 101 safe_format_map : OK : 4 {b} 101 safe_format_alt : OK : 4 {b} 101 str.format_map : OK : 4 --- 101 safe_format_map : OK : 4 {b} 101 safe_format_alt : OK : 4 {b} 101 str.format_map : OK : 4 --- 0101 safe_format_map : OK : 4 {b} 0101 safe_format_alt : OK : 4 {b} 0101 str.format_map : OK : 4 --- FOO safe_format_map : OK : 4 {b} FOO safe_format_alt : OK : 4 {b} FOO str.format_map : OK : 4 --- <built-in method index of list object at 0x7f7a485666c8> safe_format_map : OK : 4 {b} <built-in method index of list object at 0x7f7a485666c8> safe_format_alt : OK : 4 {b} <built-in method index of list object at 0x7f7a485666c8> str.format_map : OK : 4 --- Oh yes! safe_format_map : OK : 4 {b} {f[g]} safe_format_alt : OK : 4 {b} {f[g]} str.format_map : OK : 4 --- <built-in method values of dict object at 0x7f7a485da090> safe_format_map : OK : 4 {b} {f.values} safe_format_alt : OK : 4 {b} {f.values} （可能还有实际的text），但是source在我执行的大多数测试中似乎都占有优势（无论它们当然是这样）：

safe_format_map()

for text in test_texts:
    print(f'  {text}')
    %timeit safe_format(text * 1000, source)
    %timeit safe_format_alt(text * 1000, source)

Answer 19

一个非常丑陋但最简单的解决方案就是：

tmpl = '{foo}, {bar}'
tmpl.replace('{bar}', 'BAR')
Out[3]: '{foo}, BAR'

通过这种方式，您仍然可以将tmpl用作常规模板，并仅在需要时执行部分格式设置。我觉得这个问题太过微不足道，无法使用像Mohan Raj那样的过度解决方案。

Answer 20

假设你没有使用该字符串直到完全填写完毕，你可以做类似这样的事情：

class IncrementalFormatting:
    def __init__(self, string):
        self._args = []
        self._kwargs = {}
        self._string = string

    def add(self, *args, **kwargs):
        self._args.extend(args)
        self._kwargs.update(kwargs)

    def get(self):
        return self._string.format(*self._args, **self._kwargs)

示例：

template = '#{a}:{}/{}?{c}'
message = IncrementalFormatting(template)
message.add('abc')
message.add('xyz', a=24)
message.add(c='lmno')
assert message.get() == '#24:abc/xyz?lmno'

Answer 21

还有一种方法可以实现这一点，即使用format和%替换变量。例如：

>>> s = '{foo} %(bar)s'
>>> s = s.format(foo='my_foo')
>>> s
'my_foo %(bar)s'
>>> s % {'bar': 'my_bar'}
'my_foo my_bar'

Answer 22

你可以将它包装在一个带默认参数的函数中：

def print_foo_bar(foo='', bar=''):
    s = '{foo} {bar}'
    return s.format(foo=foo, bar=bar)

print_foo_bar(bar='BAR') # ' BAR'