String equivalent of +=

Question

Is there an equivalent of += for a string?

ie:

x = 1
while x <= 100:
    y = x
    if x % 3 == 0:
        y = 'Fizz'
    if x % 5 == 0:
        y += 'Buzz'
    if x % 7 == 0:
        y += 'Foo'
    if x % 11 == 0:
        y += 'Bar'
    print y
    x += 1
raw_input('Press enter to exit...')

This should return a string and a second string if the same rules as with numbers applied. Is it possible to do this? Because just doing that returns TypeError: unsupported operand type(s) for +=: 'int' and 'str', even though y is a string to begin with, not an int.

The code snippet you show is okay. Something else must be wrong. — Some programmer dude, Mar 07 '14 at 07:50
It works fine for me. Are you sure that `y` is not an integer? — Joel Cornett, Mar 07 '14 at 07:50
You can not (obviously) join a `string` with a `integer` by doing `"some string" + 1`, but you can do `"some string" + str(1)` obviously. There's no other ways of doing it. — Torxed, Mar 07 '14 at 07:52
possible duplicate of [Python String and Integer concatenation](http://stackoverflow.com/questions/2847386/python-string-and-integer-concatenation) — devnull, Mar 07 '14 at 07:52
@lc thats because the error was from another piece of code, a fizzbuzz i was making. ill edit the op with all my code — , Mar 07 '14 at 07:53
Ahh, i see what i did. If the first if is skipped, then the second assumes that the first was true, thats where the error is. — , Mar 07 '14 at 08:04
In fact a single change of `y = x` to `y = str(x)` would avoid the error. You should use `+=` for `'Fizz'` as well! — holdenweb, Jun 24 '19 at 15:59

score 3 · Answer 1 · answered Mar 07 '14 at 07:53

If you do this: You are concatenating a string to string:

x = 'a string'
x += '6'
print x

If you do this: You concatenate int to string so you get error:

x = 'a string'
x += 6
print x

error:

TypeError: cannot concatenate 'str' and 'int' objects

You have to make sure variable type before doing '+' operation; based on variable type, python can add or concatenate

score 1 · Answer 2 · edited May 11 '18 at 04:24

1

That would be s1 += s2:

>>> s1 = "a string"
>>> s1 += " and a second string"
>>> s1
'a string and a second string'
>>>

Unlike Perl, Python mostly refuses to perform implicit conversions (the numeric types being the principal exception). To concatenate the string representation of an integer i to a string s, you would have to write

s += str(i)

edited May 11 '18 at 04:24

Suren

5,775
3
26
39

answered Mar 07 '14 at 07:54

holdenweb

27,899
7
50
73

score 1 · Answer 3 · answered Mar 07 '14 at 07:55

1

The following code works for me for Python 2.7.4 and Python 3.0:

a='aaa'
a+='bbb'
print(a)
aaabbb

answered Mar 07 '14 at 07:55

vicsana1

359
1
4

score 0 · Answer 4 · edited May 23 '17 at 11:46

0

I don't know, but maybe you are looking for operator

operator.iadd(a, b)¶
operator.__iadd__(a, b)
a = iadd(a, b) is equivalent to a += b.

http://docs.python.org/2/library/operator.html

x = 'a string'
x += ' and a second string'
print x

operator.iadd(x, ' and a third string')

print x

How can I concatenate a string and a number in Python?

edited May 23 '17 at 11:46

Community

1
1

answered Mar 07 '14 at 07:55

blfuentes

2,577
4
42
65

Torxed · Answer 5 · 2014-03-07T08:19:07.243

x = 1
while x <= 100:
    y = str(x)
    if x % 3 == 0:
        y = 'Fizz'
    if x % 5 == 0:
        y += 'Buzz'
    if x % 7 == 0:
        y += 'Foo'
    if x % 11 == 0:
        y += 'Bar'
    print y
    x += 1
raw_input('Press enter to exit...')

It's quite simple. You start off by defining X as an integer, then you increese it in a while loop.
At the very beginning of each iteration you define y = x which essentially tells python to set y into an integer.

Then depending on what x modulus <nr> you got, you add a string to the integer called y (yes, it is an integer as well).. This leads to an error because it's an illegal operation because of the way a INT and a WORD works, they're just different so you need to treat one of them prior to merging them into the same variable.

How to debug your own code: Try doing print(type(x), type(y)) and you get the differences of the two variables.. Might help you wrap your head around this.

The solution is y = str(x).

So, no.. Y is NOT a string to begin with

Because you redefine y each iteration of your while loop.
y = x <-- Makes Y a int, because that's what x is :)

Also, try using .format()

x = 1
while x <= 100:
    y = x
    if x % 3 == 0:
        y = '{0}Fizz'.format(x)
    if x % 5 == 0:
        y += '{0}Buzz'.format(x)
    if x % 7 == 0:
        y += '{0}Foo'.format(x)
    if x % 11 == 0:
        y += '{0}Bar'.format(x)
    print y
    x += 1
raw_input('Press enter to exit...')

Another personal observation is that if x % 3 == 0 you replace y = ..., but in all other if-cases you append to y, why is this? I left it just the way you gave us the code but either do elif on the rest or why not concade on allif's

ahh, yes. So is this better than the solution I came up with? (look down a few answers) — , Mar 07 '14 at 08:13
@mm865 Considering i changed 1 line of code, you doubled the code.. I'd say mine makes more sense? Altho all answers are good answers i personally think you should not mess with the code to much in order to fix an issue unless you gain more functionality out of it. So in this particular case, i think my code will be understood quicker and be more efficient considering i only added 5 characters of code on 1 line in order to achieve the same outcome? :) **Also**, in 100% of the cases, you will replace `Y` with for instance `Foo`, his expected result is ` Foo`? — Torxed, Mar 07 '14 at 08:14

score 0 · Accepted Answer · answered Mar 07 '14 at 08:11

I managed to fix it using isinstance()

x = 1
while x <= 100:
    y = x
    if x % 3 == 0:
        y = 'Fizz'
    if x % 5 == 0:
        if isinstance(y, str):
            y += 'Buzz'
        else:
            y = 'Buzz'
    if x % 7 == 0:
        if isinstance(y, str):
            y += 'Foo'
        else:
            y = 'Foo'
    if x % 11 == 0:
        if isinstance(y, str):
            y += 'Bar'
        else:
           y = 'Bar'
    print y
    x += 1
raw_input('Press enter to exit...')

Please tell me if this particularly bad code (which I have a habit of writing).

@Torxed won't that mean that if the first `if` is skipped, `y` will be `xBuzz`? — , Mar 07 '14 at 08:16
@Torxed yep, tried that out just returns e.g `xFooBar` (where x is the number) instead for `FooBar` — , Mar 07 '14 at 08:19

String equivalent of +=

6 Answers6

So, no.. Y is NOT a string to begin with

Also, try using .format()