有没有Python相当于Ruby的字符串插值？

Question

Ruby示例:

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.

Answer 1

Python 3.6将添加类似于Ruby的字符串插值的文字字符串插值.从该版本的Python(计划于2016年底发布)开始,您将能够在"f-strings"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,你可以得到最接近的

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

该%运算符可用于Python中的字符串插值.第一个操作数是要插值的字符串,第二个操作数可以有不同的类型,包括"映射",将字段名称映射到要插值的值.在这里,我使用局部变量字典locals()将字段名称映射name为其值作为局部变量.

使用.format()最新Python版本方法的相同代码如下所示:

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

还有string.Template班级:

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

Answer 2

从Python 2.6.X开始,你可能想要使用:

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

Answer 3

我开发了interpy包,它可以在Python中启用字符串插值.

只需通过安装即可pip install interpy.然后,# coding: interpy在文件的开头添加行!

例:

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

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

Answer 4

Python的字符串插值类似于C的printf()

如果你试试:

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

标签%s将替换为name变量.您应该查看print函数标签:http://docs.python.org/library/functions.html

Answer 5

字符串插值将包含在PEP 498中指定的Python 3.6中.你将能够这样做:

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

请注意,我讨厌Spongebob,所以写这个有点痛苦.:)