Python 3.6及更新版本使用f-strings进行字面值字符串插值:

name='world'
print(f"Hello {name}!")

import inspect
def s(template, **kwargs):
    "Usage: s(string, **locals())"
    if not kwargs:
        frame = inspect.currentframe()
        try:
            kwargs = frame.f_back.f_locals
        finally:
            del frame
        if not kwargs:
            kwargs = globals()
    return template.format(**kwargs)

用法:

a = 123
s('{a}', locals()) # print '123'
s('{a}') # it is equal to the above statement: print '123'
s('{b}') # raise an KeyError: b variable not found

PS:性能可能是个问题。这对本地脚本有用，对生产日志没用。

复制:

Python字符串格式:% vs. .format Python中与在字符串中嵌入表达式相当的是什么?(即。“#{expr}” Ruby与Python的s= "hello， %s. "对应的是什么?%s在哪里?”%(“约翰”,“玛丽”) Python中是否有类似于Ruby的字符串插值函数?

按照PEP 498的规定，Python 3.6将包含字符串插值。你可以这样做:

name = 'Spongebob Squarepants'
print(f'Who lives in a Pineapple under the sea? \n{name}')

请注意，我讨厌海绵宝宝，所以写这篇文章有点痛苦。：）

对于旧的Python(在2.4上测试)，上面的解决方案指明了方向。你可以这样做:

import string

def try_interp():
    d = 1
    f = 1.1
    s = "s"
    print string.Template("d: $d f: $f s: $s").substitute(**locals())

try_interp()

你会得到

d: 1 f: 1.1 s: s

从Python 2.6开始。X你可能想用:

"my {0} string: {1}".format("cool", "Hello there!")

Python的字符串插值类似于C的printf()

如果你尝试:

name = "SpongeBob Squarepants"
print "Who lives in a Pineapple under the sea? %s" % name

标记%s将被name变量替换。您应该看一下打印功能标签:http://docs.python.org/library/functions.html