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。
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"))