How do I replace characters in lists using dictionary in python

Question

I want to write a code that will replace certain characters in a list in an efficient way using a dictionary.

If I have:

key = {'a':'z','b':'y','c':'x'}
List = ['a','b','c']

How can I get the output

zyx

edit to clarify. The output I want is really

randomvariableorsomething = ['z', 'y', 'x'] My apologies.

Will [key[x] for x in List] work if I don't have a key for it in the dict?

Please clarify: Your output seems to be a string. What do you mean by 'replace' in this case? — dawg
– dawg, Commented Oct 15, 2018 at 17:32
What should happen for an entry in the list with no corresponding key in the dict? ['a', 'b', 'w']? — Patrick Haugh
– Patrick Haugh, Commented Oct 15, 2018 at 17:33
Edited for clarity. Sorry, I have to get used to specifying stuff in questions. — The Overlander
– The Overlander, Commented Oct 15, 2018 at 21:15

dawg · Accepted Answer · 2018-10-15 17:40:19Z

3

Use get and join:

>>> ''.join(key.get(e,'') for e in List)
'zyx'

If by 'replace' you mean to change the list to the values of the dict in the order of the elements of the original list, you can do:

>>> List[:]=[key.get(e,'') for e in List]
>>> List
['z', 'y', 'x']

edited Oct 15, 2018 at 17:40

answered Oct 15, 2018 at 17:31

dawg

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2 Comments

Patrick Haugh Over a year ago

I was about to answer with key.get(e, e), so I suppose we should nail down what the asker wants in the case of a key that isn't in the dict.

The Overlander Over a year ago

Thanks for the help, I edited the vague question to be more accurate. Output = ['z','y','x']

Preetham · Accepted Answer · 2019-04-26 19:58:07Z

1

key = {'a':'z','b':'y','c':'x'}
List = ['a','b','c']

print([key.get(x,"No_key") for x in List])

#### Output ####

['z', 'y', 'x']

If your interest is only to print them as string,then:

print(*[key.get(x,"No_key") for x in List],sep="")

#### Output ####

zxy

Just in case you need the solution without join.

ss = ''
def fun_str(x):
    global ss
    ss = ss + x
    return(ss)
print([fun_str(x) for x in List][-1])

#### Output ####

zxy

edited Apr 26, 2019 at 19:58

answered Apr 26, 2019 at 10:34

Preetham

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3 Comments

Adrian W Over a year ago

Apparaently not what OP wanted (['z', 'y', 'x'] is not zyx), and the core of the solution is the same as the accepted answer. So, what is this good for?

Preetham Over a year ago

He has clearly mentioned as output to be in a list. And taking your comment as, feedback have edited code a bit maybe that helps.

Adrian W Over a year ago

You're right. The statement "the output I want is really" comes only after the highlighted How can I get the output zyx. Anyway, you solution is not really new - and FYI it was flagged as "low quality post". Now it is much better, I think.

G. Anderson · Accepted Answer · 2018-10-15 17:32:22Z

-1

Both keys and List are words in python that can collide with existing objects or methods (dict.keys() and List objects), so I replaced them with k and lst respectively for best practice:

[k[x] for x in lst]

edited Oct 15, 2018 at 17:32

answered Oct 15, 2018 at 17:23

G. Anderson

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2 Comments

Patrick Haugh Over a year ago

Neither of those identifiers shadows an existing identifier, nor are they keywords. keys is a method name of some classes, and typing.List exists, but neither of those effect this at all.

G. Anderson Over a year ago

You're correct, I misspoke. Edited for clarity, thanks

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