Python Iterating over a dictionary with tuples errors with sequence item 0: expected string, int found

Asked Oct 18 '15 at 07:35

Active Oct 18 '15 at 07:35

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Morning,

Bit stumped on this one.

Working in Django with Images and am trying to create a series of image sizes for a gallery. I need the value first as a tuple so that Image.thumbnail works and then as a string so I can save the file with the new size appended to it (wordpress style)

Why does the following code return a sequence item 0: expected string, int found error?

IMAGE_SIZES = {
    'THUMBNAIL':(160, 160),
    'MEDIUM':(600,600),
    'LARGE':(1000,1000),
}
for key,value in IMAGE_SIZES.items():
    temp = key + "x".join(value)
    print (temp)

It worked OK as a singular value but placing the tuple into a dictionary and iterating over it has broken it. I looked at a couple of solutions on Stack but most seem to just refer to key, value. One suggested I should join the items in the tuple but this does not seem to work with.

asked Oct 18 '15 at 07:35

Byte Insight

1

http://stackoverflow.com/questions/10880813/typeerror-sequence-item-0-expected-string-int-found – vaultah Oct 18 '15 at 07:36
tl;dr `temp = key + "x".join(map(str,value))` – Bhargav Rao Oct 18 '15 at 07:36
Dang. Almost identical to the solution that didn't work temp = 'x'.join(''.join(v) for v in value) which returned TypeError: can only join an iterable error. Thanks. – Byte Insight Oct 18 '15 at 07:40
Ah ok. I did temp = key + "x".join(str(v) for v in value) but both work – Byte Insight Oct 18 '15 at 07:40

Python Iterating over a dictionary with tuples errors with sequence item 0: expected string, int found

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