Simple syntax error in Python if else dict comprehension

Question

I have a set and dictionary and a value = 5

v = s = {'a', 'b', 'c'}
d = {'b':5 //<--new value}

If the key 'b' in dictionary d for example is in set s then I want to make that value equal to the new value when I return a dict comprehension or 0 if the key in set s is not in the dictionary d. So this is my code to do it where s['b'] = 5 and my new dictionary is is ...

{'a':0, 'b':5, 'c':0}

I wrote a dict comprehension

{  k:d[k] if k in d else k:0 for k in s}
                          ^
SyntaxError: invalid syntax

Why?! Im so furious it doesnt work. This is how you do if else in python isnt it??

So sorry everyone. For those who visited this page I originally put { k:d[k] if k in v else k:0 for k in v} and s['b'] = 5 was just a representation that the new dictionary i created would have a key 'b' equaling 5, but it isnt correct cus you cant iterate a set like that.

So to reiterate v and s are equal. They just mean vector and set.

Your description is very confused. If `s` is a set, `s['b']` doesn't make any sense. Also, what is `v` supposed to be? Is it `s`? Or something different? — abarnert, Jul 15 '13 at 22:23
Sorry it is v is supposed to be s. Im doing python vector homework — Leon, Jul 15 '13 at 22:31

vaultah · Accepted Answer · 2014-12-14T16:05:31.840

11

The expanded form of what you're trying to achieve is

a = {}
for k in v:
    a[k] = d[k] if k in d else 0

where d[k] if k in d else 0 is the Python's version of ternary operator. See? You need to drop k: from the right part of the expression:

{k: d[k] if k in d else 0 for k in v} # ≡ {k: (d[k] if k in d else 0) for k in v}

You can write it concisely like

a = {k: d.get(k, 0) for k in d}

edited Dec 14 '14 at 16:05

answered Jul 15 '13 at 22:23

vaultah

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score 3 · Answer 2 · answered Jul 15 '13 at 22:20

3

You can't use a ternary if expression for a name:value pair, because a name:value pair isn't a value.

You can use an if expression for the value or key separately, which seems to be exactly what you want here:

{k: (d[k] if k in v else 0) for k in v}

However, this is kind of silly. You're doing for k in v, so every k is in v by definition. Maybe you wanted this:

{k: (d[k] if k in v else 0) for k in d}

answered Jul 15 '13 at 22:20

abarnert

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This only seems to return me the original dictionary – Leon Jul 15 '13 at 22:47
1

@Leon: Yes, as I explained, "this is kind of silly. You're doing `for k in v`, so every `k` is in `v` by definition." This does exactly what you were trying to do in your question, which is probably not what you wanted to do, but it's hard to guess what you wanted to do without reading your mind. – abarnert Jul 15 '13 at 22:57

score 3 · Answer 3 · answered Jul 15 '13 at 22:21

3

In [82]: s = {'a', 'b', 'c'}

In [83]: d = {'b':5 }

In [85]: {key: d.get(key, 0) for key in s}
Out[85]: {'a': 0, 'b': 5, 'c': 0}

answered Jul 15 '13 at 22:21

unutbu

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Not what he wants. He's checking `k in v`, not `k in d`. And there would be no reason to do any check if `v` and `k` were the same dict, because then `key` would always be present. – abarnert Jul 15 '13 at 22:22

score 3 · Answer 4 · answered Jul 15 '13 at 22:23

3

This should solve your problem:

>>> dict((k, d.get(k, 0)) for k in s)
{'a': 0, 'c': 0, 'b': 5}

answered Jul 15 '13 at 22:23

Conan Li

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Simple syntax error in Python if else dict comprehension

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