Ruby的例子:

name = "Spongebob Squarepants"
puts "Who lives in a Pineapple under the sea? \n#{name}."

成功的Python字符串连接对我来说似乎很冗长。


当前回答

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。:)

其他回答

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。:)

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

Python 3.6将添加文字字符串插值,类似于Ruby的字符串插值。从该版本的Python开始(计划于2016年底发布),您将能够在“f-字符串”中包含表达式,例如。

name = "Spongebob Squarepants"
print(f"Who lives in a Pineapple under the sea? {name}.")

在3.6之前,你能得到的最接近的结果是

name = "Spongebob Squarepants"
print("Who lives in a Pineapple under the sea? %(name)s." % locals())

在Python中,%操作符可用于字符串插值。第一个操作数是要插入的字符串,第二个操作数可以有不同的类型,包括“mapping”,将字段名映射到要插入的值。在这里,我使用局部变量字典locals()将字段名name映射到其值作为局部变量。

使用最新Python版本的.format()方法的相同代码看起来像这样:

name = "Spongebob Squarepants"
print("Who lives in a Pineapple under the sea? {name!s}.".format(**locals()))

还有字符串。模板类:

tmpl = string.Template("Who lives in a Pineapple under the sea? $name.")
print(tmpl.substitute(name="Spongebob Squarepants"))