如何在Python中创建匿名函数而不对使用Christening ?

  • 回答 (2)
  • 关注 (0)
  • 查看 (59)

是否可以将函数放在数据结构中,而不首先给它命名def?

# This is the behaviour I want. Prints "hi".
def myprint(msg):
    print msg
f_list = [ myprint ]
f_list[0]('hi')
# The word "myprint" is never used again. Why litter the namespace with it?

lambda函数的身体非常有限,所以我不能使用它们。

def handle_message( msg ):
    print msg
def handle_warning( msg ):
    global num_warnings, num_fatals
    num_warnings += 1
    if ( is_fatal( msg ) ):
        num_fatals += 1
handlers = (
    ( re.compile( '^<\w+> (.*)' ), handle_message ),
    ( re.compile( '^\*{3} (.*)' ), handle_warning ),
)
# There are really 10 or so handlers, of similar length.
# The regexps are uncomfortably separated from the handler bodies,
# and the code is unnecessarily long.

for line in open( "log" ):
    for ( regex, handler ) in handlers:
        m = regex.search( line )
        if ( m ): handler( m.group(1) )
名字真难选a名字真难选a提问于
活动执行wrl回答于
已采纳
import re

# List of pairs (regexp, handler)
handlers = []

def handler_for(regexp):
    """Declare a function as handler for a regular expression."""
    def gethandler(f):
        handlers.append((re.compile(regexp), f))
        return f
    return gethandler

@handler_for(r'^<\w+> (.*)')
def handle_message(msg):
    print msg

@handler_for(r'^\*{3} (.*)')
def handle_warning(msg):
    global num_warnings, num_fatals
    num_warnings += 1
    if is_fatal(msg):
        num_fatals += 1
hyper_smurf2012软件工程师回答于
def message(msg):
    print msg
message.re = '^<\w+> (.*)'

def warning(msg):
    global num_warnings, num_fatals
    num_warnings += 1
    if ( is_fatal( msg ) ):
        num_fatals += 1
warning.re = '^\*{3} (.*)'

handlers = [(re.compile(x.re), x) for x in [
        message,
        warning,
        foo,
        bar,
        baz,
    ]]

扫码关注云+社区

领取腾讯云代金券