语法练习：parrot_trouble

语法练习：parrot_trouble

题目：parrot_trouble

We have a loud talking parrot. The “hour” parameter is the current hour time in the range 0…23. We are in trouble if the parrot is talking and the hour is before 7 or after 20. Return True if we are in trouble.

parrot_trouble(True, 6) → True
parrot_trouble(True, 7) → False
parrot_trouble(False, 6) → False

我的解答：

def parrot_trouble(talking, hour):
  if talking&(hour < 7 or hour > 20):
    return True
  else:
    return False
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Expected Run
parrot_trouble(True, 6) → True True OK
parrot_trouble(True, 7) → False False OK
parrot_trouble(False, 6) → False False OK
parrot_trouble(True, 21) → True True OK
parrot_trouble(False, 21) → False False OK
parrot_trouble(False, 20) → False False OK
parrot_trouble(True, 23) → True True OK
parrot_trouble(False, 23) → False False OK
parrot_trouble(True, 20) → False False OK
parrot_trouble(False, 12) → False False OK

All Correct

标答：

def parrot_trouble(talking, hour):
  return (talking and (hour < 7 or hour > 20))
  # Need extra parenthesis around the or clause
  # since and binds more tightly than or.
  # and is like arithmetic *, or is like arithmetic +
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