CODEWITHSUNDEEP
pythonpython-3.xdictionaryif-statementvalueerror
AuthorDesigned
AuthorDesigned

Reputation: 23

Python: If/elif/else Statement Not Working as Expected?

So I have the user input an expression:

Example:

1 + 1
d * 3
100 / j
s * c

To allow the user to mix numbers (int) and letters (str) I have the following if statement:

user_input = input("Enter a math Equation: ")
user_input = user_input.strip().split(" ")
user_c1 = user_input[0]
user_c2 = user_input[2]
user_c3 = user_input[1]
l1 = user_c1
l2 = user_c2
l3 = user_c2
l6 = user_c1
l4 = int(user_c2)
l5 = int(user_c1)
l7 = int(user_c1)
l8 = int(user_c2)


if l1 and l2 in kb_dxnry:
    print("1st")
elif l3 in kb_dxnry and l4 not in kb_dxnry:
    print("2nd")
elif l5 not in kb_dxnry and l6 in kb_dxnry:
    print("3rd")
else:
    print("4th")

Dictionary Code

user_kb_dxnry = {'a': 1, 'b': 2, 'c': 3, 'd': 4, 'e': 5, 'f': 6,
                 'g': 7, 'h': 8, 'i': 9, 'j': 10, "k": 11, "l": 12,
                 "m": 13, "n": 14, "o": 15, "p": 16, "q": 17, "r": 18,
                 "s": 19, 't': 20, 'u': 21, 'v': 22, 'w': 23, 'x': 24,
                 'y': 25, 'z': 26, "0": 0
                 }
kb_dxnry = user_kb_dxnry

If I type (well testing it): 1 + 1 it will execute the else statement (as it should). However, if you type any of the other combinations:

a + a
1 + a
a + 1

It ignores the if and elif statements and tries to executes else & failes.

So I type

23 * 23

Output:

4th

I type

a + 1 or a + a or 1 * a

Output is:

ValueError: invalid literal for int() with base 10:

So my if statement is failing and being new to (programming in general) I was wondering if there is a better way to do what I'm trying to do above. Maybe something I've over looked or under looked?

Would a for be better suited or would it give me the same error?

Feel free to link me to any thing you think could be helpful in helping me get this working.

EDIT:

Perhaps this would be better:

user_input = input("Enter a math Equation: ")
    user_input = user_input.strip().split(" ")
    user_c1 = user_input[0]
    user_c2 = user_input[2]
    user_c3 = user_input[1]
    l1 = user_c1
    l2 = user_c2
    l3 = user_c2
    l6 = user_c1
   user_input = input("Enter a math Equation: ")
user_input = user_input.strip().split(" ")
user_c1 = user_input[0]
user_c2 = user_input[2]
user_c3 = user_input[1]
l1 = user_c1
l2 = user_c2
l3 = user_c2
l6 = user_c1
l4 = user_c2
l5 = user_c1
l7 = user_c1
l8 = user_c2


if l1 in kb_dxnry and l2 in kb_dxnry:
    print("1st")
elif l3 in kb_dxnry and l4 not in kb_dxnry:
    l4 = int(user_c2)
    print("2nd")
elif l5 not in kb_dxnry and l6 in kb_dxnry:
    l5 = int(user_c1)
    print("3rd")
else:
    l7 = int(user_c1)
    l8 = int(user_c2)
    print("4th")

Alright with the rewrite of code above the ErrorValue is gone. But if you type a + 1 or 1 + a

It still skips elif 1 & 2:

elif l3 in kb_dxnry and l4 not in kb_dxnry:
    print("2nd")
elif l6 in kb_dxnry and l5 not in kb_dxnry:
    print("3rd")

And jumps to else.

Upvotes: 0

Views: 172

Answers (2)

AuthorDesigned
AuthorDesigned

Reputation: 23

Alright I figured it out with the help of @AirSquid. I removed the str to int and search for the strings that may be numbers in the dictionary. Than in the if statement is will I will change them to int.

user_input = input("Enter a math Equation: ")
    user_input = user_input.strip().split(" ")
    user_c1 = user_input[0]
    user_c2 = user_input[2]
    user_c3 = user_input[1]
    l1 = user_c1
    l2 = user_c2
    l3 = user_c2
    l6 = user_c1
   user_input = input("Enter a math Equation: ")
user_input = user_input.strip().split(" ")
user_c1 = user_input[0]
user_c2 = user_input[2]
user_c3 = user_input[1]
l1 = user_c1
l2 = user_c2
l3 = user_c2
l6 = user_c1
l4 = user_c2
l5 = user_c1
l7 = user_c1
l8 = user_c2


if l1 in kb_dxnry and l2 in kb_dxnry:
    print("1st")
elif l3 in kb_dxnry:
    l4 = int(user_c2)
    print("2nd")
elif l5 not in kb_dxnry:
    l5 = int(user_c1)
    print("3rd")
else:
    l7 = int(user_c1)
    l8 = int(user_c2)
    print("4th")

The above code now does what I wanted it to do.

Upvotes: 0

AirSquid
AirSquid

Reputation: 11903

I'm not sure why you are making so many variables if you are only trying to process one expression.

If you look at the line number in your error statement, I bet you will see it refers to the lines where you are defining l4, l5, l7, l7. You are doing int() conversions there and if you pass those conversions a string, you will get the error you posted, so I don't think you are even getting to your if-else statement.

Upvotes: 1

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