关于python的type函数问题

题目：Assume that two variables, varA and varB, are assigned values, either numbers or strings.
Write a piece of Python code that prints out one of the following messages:
"string involved" if either varA or varB are strings
"bigger" if varA is larger than varB
"equal" if varA is equal to varB
"smaller" if varA is smaller than varB

我这样写
if type(varA) or type(varB) ==type< 'str'>:print('string involved')
if varA > varB:print('bigger')
elif varA == varB:print('equal')
elif varA < varB:print('smaller')

总是第一句不对。它说：
File "submission.py", line 1
if type(varA) or type(varB) ==type< 'str'>:print('string involved')
^
SyntaxError: invalid syntax

我不明白……

if type(varA) == type("") or type(varB) == type(""):print('string involved')

你拼错了，python里的构造函数应该是 def __init__()而不是 def _init_()请采纳

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