NorthIsUp NorthIsUp - 1 year ago 64
Python Question

Python string formatting: % vs. .format

Python 2.6 introduced the

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?

    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 Source

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().