`SyntaxError` in if-else one-liner

Question

I wrote this block of code, but it doesn't work because of SyntaxError.

def char_freq(message):
    d = dict()
    for ch in message:
        d[ch] += 1 if ch in d else d[ch] = 1
    return d                             ^ SyntaxError: End of statement expected

I don't know how to rewrite the expression in order to keep if-else in one line and to get it to work.

I know it is possible to implement the function as a simple for loop, but I don't understand, why my if-else one-liner results in SyntaxError?

Thanks, it works. But I don't understand why notation in my question doesn't work. — Evgeniy Golovin, Jan 08 '21 at 18:11

Sayse · Answer 1 · 2021-01-08T11:21:59.013

2

Turn d into a defaultdict and then you can just ignore the statement altogether

from collections import defaultdict
def char_freq(message):
    d = defaultdict(int)
    for ch in message:
        d[ch] += 1
    return d

It also looks like you're just counting characters so you could just use a counter

from collections import Counter
def char_freq(message):
    return Counter(message)

edited Jan 08 '21 at 11:21

answered Jan 08 '21 at 11:15

Sayse

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1

Thanks for reply, great ways to improve my code! – Evgeniy Golovin Jan 08 '21 at 18:17

score 1 · Accepted Answer · answered Jan 08 '21 at 11:12

1

As you asked to keep the if/else Do

d[ch] = (d[ch] + 1) if ch in d else 1

But the dict.get() syntax is nicer d[ch] = d.get(ch, 0) + 1

Or a collections.defaultdict with int factory

def char_freq(message):
    d = defaultdict(int)
    for ch in message:
        d[ch] += 1
    return d

answered Jan 08 '21 at 11:12

azro

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Thanks for the answer. I'm still curious, why my notation results in error? – Evgeniy Golovin Jan 08 '21 at 18:15
@EvgeniyGolovin you can't do assignement in if/else parts – azro Jan 08 '21 at 18:23

`SyntaxError` in if-else one-liner

2 Answers2