是否可以使用类似于字符串模板safe_substitute()功能的高级字符串格式化方法进行部分字符串格式化？
例如：
s = '{foo} {bar}'
s.format(foo='FOO') #Problem: raises KeyError 'bar'
问问题
33272 次
23 回答
150
如果您知道格式化内容的顺序：
s = '{foo} {{bar}}'
像这样使用它：
ss = s.format(foo='FOO')
print ss
>>> 'FOO {bar}'
print ss.format(bar='BAR')
>>> 'FOO BAR'
您不能同时指定foo和bar- 您必须按顺序执行。
于 2013-08-20T19:39:51.260 回答
141
您可以使用简短、最易读并且还描述了编码器意图的partial函数：functools
from functools import partial
s = partial("{foo} {bar}".format, foo="FOO")
print s(bar="BAR")
# FOO BAR
于 2016-12-13T00:15:41.070 回答
67
您可以通过覆盖映射将其欺骗为部分格式：
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，这个基本实现只适用于基本情况。
于 2012-07-01T17:07:17.783 回答
58
这种限制.format()——无法进行部分替换——一直困扰着我。
在评估了Formatter此处许多答案中描述的自定义类，甚至考虑使用第三方包（如lazy_format ）后，我发现了一个更简单的内置解决方案：模板字符串
它提供了类似的功能，但也提供了部分替换彻底safe_substitute()的方法。模板字符串需要有一个$前缀（感觉有点奇怪——但我认为整体解决方案更好）。
import string
template = string.Template('${x} ${y}')
try:
template.substitute({'x':1}) # raises KeyError
except KeyError:
pass
# but the following raises no error
partial_str = template.safe_substitute({'x':1}) # no error
# partial_str now contains a string with partial substitution
partial_template = string.Template(partial_str)
substituted_str = partial_template.safe_substitute({'y':2}) # no error
print substituted_str # prints '12'
基于此形成了一个方便的包装器：
class StringTemplate(object):
def __init__(self, template):
self.template = string.Template(template)
self.partial_substituted_str = None
def __repr__(self):
return self.template.safe_substitute()
def format(self, *args, **kws):
self.partial_substituted_str = self.template.safe_substitute(*args, **kws)
self.template = string.Template(self.partial_substituted_str)
return self.__repr__()
>>> s = StringTemplate('${x}${y}')
>>> s
'${x}${y}'
>>> s.format(x=1)
'1${y}'
>>> s.format({'y':2})
'12'
>>> print s
12
同样，基于 Sven 的回答的包装器使用默认字符串格式：
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)
于 2014-04-26T01:25:42.393 回答
29
不确定这是否可以作为一种快速解决方法，但是怎么样
s = '{foo} {bar}'
s.format(foo='FOO', bar='{bar}')
? :)
于 2015-04-07T11:07:26.897 回答
11
如果您定义自己的Formatter覆盖该方法的get_value方法，则可以使用它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，您可以映射bar到"{bar}"if baris not in the kwargs。
但是，这需要使用format()Formatter 对象的方法，而不是字符串的format()方法。
于 2012-07-01T17:06:33.093 回答
10
>>> 'fd:{uid}:{{topic_id}}'.format(uid=123)
'fd:123:{topic_id}'
试试这个。
于 2015-08-17T09:19:39.303 回答
7
感谢Amber的评论，我想出了这个：
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
于 2013-03-31T09:57:51.230 回答
4
我发现的所有解决方案似乎都存在更高级规格或转换选项的问题。@SvenMarnach 的FormatPlaceholder非常聪明，但它不能与强制（例如{a!s:>2s}）一起正常工作，因为它调用__str__方法（在这个例子中）而不是__format__你失去任何额外的格式。
这是我最终得到的结果以及其中的一些关键功能：
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（不仅仅是映射）类似的接口 - 支持更复杂的格式化选项：
- 强迫
{k!s}{!r} - 嵌套
{k:>{size}} - 获取属性
{k.foo} - 获取项目
{k[0]} - 强制+格式化
{k!s:>{size}}
- 强迫
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表现的测试之后，我发现了各种实现的问题。如果有人发现它们很有见地，它们就在下面。
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)
于 2020-04-08T09:12:24.967 回答
3
对我来说这已经足够好了：
>>> 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'
于 2016-03-03T15:06:22.647 回答
1
一个非常丑陋但对我来说最简单的解决方案就是：
tmpl = '{foo}, {bar}'
tmpl.replace('{bar}', 'BAR')
Out[3]: '{foo}, BAR'
这样，您仍然可以将tmpl其用作常规模板并仅在需要时执行部分格式化。我发现这个问题太简单了，无法使用像 Mohan Raj 这样的过度杀伤性解决方案。
于 2018-04-24T09:48:59.330 回答
1
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：这里显示了
一种更优雅的方式（子类化dict和重载__missing__(self, key)）：https ://stackoverflow.com/a/17215533/333403
于 2019-06-07T08:39:36.340 回答
1
如果您想解压缩字典以将参数传递给format，就像在这个相关问题中一样，您可以使用以下方法。
首先假设字符串s与此问题中的相同：
s = '{foo} {bar}'
并且值由以下字典给出：
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，您可以首先从其中获取set所有命名参数s并创建一个字典，将参数映射到用花括号括起来的自身：
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字典来填充replacements. 对于 python 3.5+，您可以在单个表达式中执行此操作：
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}'
于 2019-11-25T16:24:50.127 回答
1
这是一个基于正则表达式的轻度hacky 解决方案。请注意，这不适用于嵌套格式说明符，例如{foo:{width}}，但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'
于 2020-09-16T16:13:05.567 回答
0
还有另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'
于 2016-08-28T22:16:41.583 回答
0
假设您在完全填写之前不会使用字符串，您可以执行类似此类的操作：
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)
例子：
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'
于 2017-05-22T18:16:43.330 回答
0
在从这里和那里测试了最有希望的解决方案之后，我意识到它们都没有真正满足以下要求：
- 严格遵守模板认可的语法
str.format_map()； - 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，满足上述要求。（编辑：现在@SvenMarnach 的版本——正如这个答案中所报告的——似乎可以处理我需要的极端情况）。
基本上，我最终解析了模板字符串，找到匹配的嵌套{.*?}组（使用find_all()辅助函数）并逐步构建格式化字符串，并直接使用str.format_map()同时捕获任何潜在KeyError的 .
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))
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
（此代码也可在FlyingCircus中找到——免责声明：我是它的主要作者。）
此代码的用法是：
print(safe_format_map('{a} {b} {c}', dict(a=-A-)))
# -A- {b} {c}
让我们将此与我最喜欢的解决方案（@SvenMarnach在这里和那里分享他的代码）进行比较：
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))
这里有几个测试：
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=[])
以及使其运行的代码：
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}')
导致：
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}
如您所见，更新版本现在似乎可以很好地处理早期版本失败的极端情况。
从时间上看，它们在大约。彼此的 50%，取决于实际text的格式（可能是实际的source），但safe_format_map()在我执行的大多数测试中似乎都有优势（当然，不管它们是什么意思）：
for text in test_texts:
print(f' {text}')
%timeit safe_format(text * 1000, source)
%timeit safe_format_alt(text * 1000, source)
{b} {f}
3.93 ms ± 153 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
6.35 ms ± 51.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b}
4.37 ms ± 57.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
5.2 ms ± 159 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {c:5d}
7.15 ms ± 91.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
7.76 ms ± 69.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {c!s}
7.04 ms ± 138 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
7.56 ms ± 161 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {c!s:>{a}s}
8.91 ms ± 113 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
10.5 ms ± 181 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {c:0{a}d}
8.84 ms ± 147 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
10.2 ms ± 202 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {d[x]}
7.01 ms ± 197 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
7.35 ms ± 106 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {e.index}
11 ms ± 68.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
8.78 ms ± 405 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {f[g]}
6.55 ms ± 88.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
9.12 ms ± 159 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
{a} {b} {f.values}
6.61 ms ± 55.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
9.92 ms ± 98.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
于 2019-07-26T18:32:39.280 回答
0
我喜欢@sven-marnach 的回答。我的回答只是它的扩展版本。它允许非关键字格式并忽略额外的键。以下是使用示例（函数名称是对 python 3.6 f-string 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)
于 2020-02-04T13:40:58.700 回答
0
如果您正在做大量的模板化并发现 Python 的内置字符串模板功能不足或笨拙，请查看Jinja2。
从文档：
Jinja 是一种现代且对设计人员友好的 Python 模板语言，以 Django 的模板为模型。
于 2020-03-12T15:30:58.657 回答
0
阅读@Sam Bourne 的评论，我修改了@SvenMarnach 的代码{a!s:>2s}以在不编写自定义解析器的情况下
使用强制（如）正常工作。基本思想不是转换为字符串，而是将缺少的键与强制标签连接起来。
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 的测试启发）在两种情况下检查传统format_map和safe_format_map基于上述类的输出：提供正确的关键字或不提供正确的关键字。
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>
于 2020-07-03T16:33:49.187 回答
0
TL; DR：问题：如果未设置则defaultdict失败：{foobar[a]}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
DefaultWrapper解决方案：从Edit复制类，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。以下代码引发了一个TypeError: string indices must be integers.
from collections import defaultdict
text = "{foo} '{bar[index]}'"
text.format_map(defaultdict(str, foo="FOO")) # raises a TypeError
为了解决这个问题，可以将该collections.defaultdict解决方案与支持索引的自定义默认值对象一起使用。对象在索引和属性访问时返回自身，DefaultWrapper允许无限次索引/使用属性而不会出错。
请注意，这可以扩展为允许包含部分请求值的容器。看看下面的编辑。
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：部分装满的容器
上述类可以扩展为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defaultdict用DefaultWrapper. DefaultWrapper对象环绕返回的container容器请求值（用DefaultWrapper对象包裹）或容器作为字符串。这样可以模拟无限深度的地图，但会返回所有当前值。
kwargs仅为了方便而添加。这样它看起来更像是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可以正常工作：
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', ''
于 2021-02-01T10:43:52.093 回答
0
这就是我们设法做到这一点的方法：
import traceback
def grab_key_from_exc(exc):
last_line_idx = exc[:-1].rfind('\n')
last_line = exc[last_line_idx:]
quote_start = last_line.find("'")
quote_end = last_line.rfind("'")
key_name = last_line[quote_start+1:quote_end]
return key_name
def partial_format(input_string, **kwargs):
while True:
try:
return input_string.format(**kwargs)
except:
exc = traceback.format_exc()
key = grab_key_from_exc(exc)
kwargs[key] = '{'+key+'}'
于 2021-12-13T10:43:52.577 回答
-3
您可以将其包装在一个采用默认参数的函数中：
def print_foo_bar(foo='', bar=''):
s = '{foo} {bar}'
return s.format(foo=foo, bar=bar)
print_foo_bar(bar='BAR') # ' BAR'
于 2012-07-01T17:07:48.553 回答