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

我开发了interpy包,在Python中支持字符串插值。

只需通过pip install interpy安装即可。 然后,在文件的开头添加一行# coding: interpy !

例子:

#!/usr/bin/env python
# coding: interpy

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

其他回答

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}!")

你也可以吃这个

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

http://docs.python.org/2/library/string.html#formatstrings

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

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。:)

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"))