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martinhugh
Group Title
Question:
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
Answer I tried:
if (type(varA) or type(varB))=='str':
print ("string involved")
else:
if varA>varB:
print("bigger")
if varA=varB:
print("equal")
if varA<varB:
print("smaller")
Shows wrong
 one year ago
 one year ago
martinhugh Group Title
Question: 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 Answer I tried: if (type(varA) or type(varB))=='str': print ("string involved") else: if varA>varB: print("bigger") if varA=varB: print("equal") if varA<varB: print("smaller") Shows wrong
 one year ago
 one year ago

This Question is Open

tyteen4a03 Group TitleBest ResponseYou've already chosen the best response.1
The problems comes from this statement: if varA=varB: You cannot use the = sign to compare variables, the = sign in Python (and any other languages) is used to assign values. Use == for comparison, and === for typesensitive comparison. Also, the statement if (type(varA) or type(varB))=='str': is wrong. The or operator is used for logic, so (type(varA) or type(varB)) will return True (because type(varA/B) will return either str or int) and the whole statement will return to False (because True != "str"). There are 2 ways to do this, one is: if "str" in [type(varA), type(varB)]. This statement creates a temporary list and checks if either of them is a string. Another way to do this is: if type(varA) == "str" or type(varB) == "str". This is the correct way of showing the logic, but less elegant in my opinion.
 one year ago

condensermike Group TitleBest ResponseYou've already chosen the best response.0
In addition, you are not using 'elif' in your code. You have an 'else:' followed by three 'if' statements. This is not correct. Review this portion in the notes. Lastly, in the equal to comparison you are using a single = sign. Go back and review this. A single equal sign is only used for assignment. Hope this helps!
 one year ago

condensermike Group TitleBest ResponseYou've already chosen the best response.0
Also, you have omitted the parentheses in your conditional statements, you must have these. ex: if(a > b) .....
 one year ago

msmithhnova Group TitleBest ResponseYou've already chosen the best response.0
I got the above code to work making only 2 changes. If varA=varB changed to If varA==varB < two == for equality and == 'str' changed to ==type('') <compares to type known to be string
 one year ago

msmithhnova Group TitleBest ResponseYou've already chosen the best response.0
BTW, the last part type('') just has two apostrophes in the brackets (empty string).
 one year ago
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