考虑这个虚构的 Python 函数:
def f(s):
# accepts a string containing placeholders
# returns an interpolated string
return s % {'foo': 'OK', 'bar': 'OK'}
如何检查字符串s是否提供了所有预期的占位符,如果没有,使函数礼貌地显示缺少的键?
我的解决方案如下。我的问题:有更好的解决方案吗?
import sys
def f(s):
d = {}
notfound = []
expected = ['foo', 'bar']
while True:
try:
s % d
break
except KeyError as e:
key = e.args[0] # missing key
notfound.append(key)
d.update({key: None})
missing = set(expected).difference(set(notfound))
if missing:
sys.exit("missing keys: %s" % ", ".join(list(missing)))
return s % {'foo': 'OK', 'bar': 'OK'}