Wednesday, 10 May 2017

python - String formatting: % vs. .format



Python 2.6 introduced the str.format() method with a slightly different syntax from the existing % operator. Which is better and for what situations?




  1. The following uses each method and has the same outcome, so what is the difference?



    #!/usr/bin/python

    sub1 = "python string!"
    sub2 = "an arg"

    a = "i am a %s" % sub1
    b = "i am a {0}".format(sub1)

    c = "with %(kwarg)s!" % {'kwarg':sub2}
    d = "with {kwarg}!".format(kwarg=sub2)

    print a # "i am a python string!"

    print b # "i am a python string!"
    print c # "with an arg!"
    print d # "with an arg!"

  2. Furthermore when does string formatting occur in Python? For example, if my logging level is set to HIGH will I still take a hit for performing the following % operation? And if so, is there a way to avoid this?



    log.debug("some debug info: %s" % some_info)


Answer




To answer your first question... .format just seems more sophisticated in many ways. An annoying thing about % is also how it can either take a variable or a tuple. You'd think the following would always work:



"hi there %s" % name


yet, if name happens to be (1, 2, 3), it will throw a TypeError. To guarantee that it always prints, you'd need to do



"hi there %s" % (name,)   # supply the single argument as a single-item tuple



which is just ugly. .format doesn't have those issues. Also in the second example you gave, the .format example is much cleaner looking.



Why would you not use it?




  • not knowing about it (me before reading this)

  • having to be compatible with Python 2.5







To answer your second question, string formatting happens at the same time as any other operation - when the string formatting expression is evaluated. And Python, not being a lazy language, evaluates expressions before calling functions, so in your log.debug example, the expression "some debug info: %s"%some_infowill first evaluate to, e.g. "some debug info: roflcopters are active", then that string will be passed to log.debug().


No comments:

Post a Comment

c++ - Does curly brackets matter for empty constructor?

Those brackets declare an empty, inline constructor. In that case, with them, the constructor does exist, it merely does nothing more than t...