Invalid syntax with if

Question

this is a pretty simple idea, you enter your test score and if you got above a 70% (35/50) you can do corrections for 1 pt back essentially giving you a 100%. If you get under 70% you can do corrections for 1/2 a point back.

this is giving me a invalid syntax and putting the cursor between the last " and )

score = input("How many problems did you get right on the test?")
maxscore = 50
passscore = 35
wrong = (maxscore - score)

if (score > passscore):
    print ("You will get a 100%")


if (score < passscore):
    print("You can get"(wrong)"% back  with text corrections")

Im terrible at programing so sorry if i seem really stupid here.

You need to append `wrong` to the other two strings on each side of it. Using + or ,. E.g. `print("Hello" + "World") ` — Luke Joshua Park, Oct 13 '15 at 22:25
Related (not exactly a dupe though): http://stackoverflow.com/questions/13945749/string-formatting-in-python-3 — Daniel Pryden, Oct 13 '15 at 22:25
@TigerhawkT3 Because `print ("You will get a 100%")` and `input()` function , I think OP is using Python 3. — Remi Guan, Oct 13 '15 at 22:32
What if score is equal to passscore? Plus, how do you feel wrong is a percentage? — Bill Woodger, Oct 13 '15 at 22:34
@KevinGuan - Technically, you can use parentheses in a Python 2 `print` statement. Something like `print ('hi')` produces identical results in Python 2 and 3, but `print ('hi', 'there')` produces the tuple `('hi', 'there')` in Python 2 and the string `'hi there'` in Python 3. Also, `input()` exists in Python 2, and will evaluate the entered string into an `int` or `float` or whatever - it's Python 3 that would need to cast the string. — TigerhawkT3, Oct 13 '15 at 22:41
@TigerhawkT3 However usually we use `print 'foobar'` in Python 2 :) And about `print ('hi')` can works on Python 2 because it will create a tuple that only one element in it so the `()` willn't be print out? — Remi Guan, Oct 13 '15 at 22:45
@KevinGuan - Nope, that's not a `tuple`. The parentheses only affect grouping. `1,` is a `tuple`, while `(1)` is an `int`. — TigerhawkT3, Oct 13 '15 at 23:10

score 3 · Accepted Answer · answered Oct 13 '15 at 22:27

The problem is here:

print("You can get"(wrong)"% back  with text corrections")

This is not the correct way to insert a variable into a string. You have several options:

print("You can get " + str(wrong) + "% back with text corrections")

Or:

print("You can get %d%% back with text corrections" % wrong)

Or:

print("You can get {}% back with text corrections".format(wrong))

Or:

print("You can get ", wrong, "% back with text corrections", sep='')

Also, if you're using Python 3, you'll need to do score = int(input(... to cast the string you received as an integer.

score 2 · Answer 2 · answered Oct 13 '15 at 22:49

Everyone has to start somewhere (and I'm still pretty new to Python myself)!

Your first problem is that you need to define score as an int:

score = int(input("How many problems did you get right on the test?"))

Then there are at least two solutions to fix that last line of code. One is to use + to separate your strings of text, plus str to convert wrong to string format:

print("You can get " + str(wrong) + "% back  with text corrections")

Or you can use the .format approach, which is more "Pythonic":

print("You can get {0}% back  with text corrections".format(wrong))

score 1 · Answer 3 · answered Oct 13 '15 at 22:25

1

You can put multiple arguments by comma separating them..

print "You can get", wrong, "% back  with text corrections"

answered Oct 13 '15 at 22:25

ergonaut

6,929
1
17
47

Dimitris Fasarakis Hilliard · Answer 4 · 2015-10-13T23:50:25.860

Before you worry about the syntactic error, you need to transform wrong to an int because input() returns user input as a str type.

score = int(input("How many problems did you get right on the test?"))

If you don't and the user enters a string, the expression:

wrong = (maxscore - score)

Will raise a TypeError, which essentially means that you cannot subtract a value of type str (score) from a value of type int (maxscore).

As for your syntactic error.

print("You can get"(wrong)"% back  with text corrections")

is syntactically invalid. You need to include wrong as a string by transforming it with str() in your print() call:

print("You can get" + str(wrong) + "% back  with text corrections")

You can see, conversions between different types, depending on the operation can be a mess until you get a hang of them.

digitaLink · Answer 5 · 2015-10-13T22:32:15.803

1

If you want a string concatenated you need to add + between variable names and Strings. Replace your second print line with:

print("You can get " + str(wrong) + "% back  with text corrections")

edited Oct 13 '15 at 22:32

answered Oct 13 '15 at 22:26

digitaLink

458
3
17

score 1 · Answer 6 · answered Oct 13 '15 at 22:29

First, score must be int. So you need use int() function to do that.
And if don't need (), just remove them.
Then , the problem is about print("You can get"(wrong)"% back with text corrections"), you should use + or , or .format(), etc. here. And remember use str() to convert it to string.

score = int(input("How many problems did you get right on the test?"))
maxscore = 50
passscore = 35
wrong = (maxscore - score)

if score > passscore:
    print("You will get a 100%")


if score < passscore:
    print("You can get "+str(wrong)+"% back with text corrections")

This is the simplest way, but use .format() will be more clear like this:

print("You can get {0}% back  with text corrections".format(wrong))

Or like this:

print("You can get {wrong}% back  with text corrections".format(wrong=wrong))

Invalid syntax with if

6 Answers6