Replacing symbols using list comprehension

Question

I want to simplify replacing specific characters of a string in-situ - with a list comprehension. Attempts so far simply return a list of strings - each list item with each character replaced from the check string.

Advice / solutions?

Inputs:

reveal = "password"
ltrTried = "sr"

Required Output:

return = "**ss**r*"

Getting:

('**ss****', '******r*')

List comprehensions aren't meant for situations with side effects, but what you're looking for here is entirely a side effect (since you aren't generating a list you want to keep). What is wrong with your for loop, and why do you want to replace it with a comprehension? — David Robinson, Sep 09 '14 at 23:11
See [here](http://stackoverflow.com/questions/5753597/is-it-pythonic-to-use-list-comprehensions-for-just-side-effects), [here](http://stackoverflow.com/questions/8068251/why-is-python-list-comprehension-sometimes-frowned-upon) and [here](http://stackoverflow.com/questions/8695488/proper-use-of-list-comprehensions-python) for more on list comprehension side effects. — David Robinson, Sep 09 '14 at 23:11
As a side note, they're "list comprehensions", not "comprehension lists". It's not a special kind of list, it's a perfectly ordinary one, just built in a special way. — abarnert, Sep 10 '14 at 00:04

score 2 · Accepted Answer · answered Sep 09 '14 at 23:19

2

If you want to do this using a list comprehension, you'd want to replace it letter by letter like this:

reveal = "".join((letter if letter in ltrFound else "*") for letter in reveal)

Notice that

We're iterating over your reveal string, not your ltrFound list (or string).
Each item is replaced using the ternary operator letter if letter in ltrFound else "*". This ensures that if the letter in reveal is not in ltrFound, it will get replaced with a *.
We end by joining together all the letters.

answered Sep 09 '14 at 23:19

David Robinson

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That's what I was missing.... ELSE. Thanks for putting me out of my misery. Probably obvious I've only been using Python for a few weeks. Wanted to explore comprehension lists - but couldn't see how to get it to work this way around. Replacing the letters I had was easy - replacing the ones I didn't was driving me nuts! – alcyone_pleiades Sep 09 '14 at 23:31
this is a generator not a list comprehension – Padraic Cunningham Sep 09 '14 at 23:37
@alcyone_pleiades: The `if` here is different from the `if` in your original code. You were using an `if` clause in a comprehension, which filters the values, only iterating the ones for which the `if` is true. This answer is using an `if-else` expression, which always evaluates to a value, but one of two different values depending on whether the test is true. They may look similar, but don't mix them up. – abarnert Sep 10 '14 at 00:01
@PadraicCunningham: Good point. A generator expression is a kind of comprehension, although a different kind than a list comprehension of course. In this case, it's clearer to use a generator expression because we don't actually need the list for anything, but it's probably worth explaining that in the answer rather than just changing it without comment. – abarnert Sep 10 '14 at 00:02
@abarnet - no confusion here - I know they're different. Just wasn't aware you could place ELSE here. – alcyone_pleiades Sep 10 '14 at 00:45
@alcyone_pleiades: If that's true, I'm pretty sure you _are_ confused. You _have_ to have an `else` in a conditional expression. So whatever you were thinking of that you didn't think could have an `else`, that's not it. – abarnert Sep 10 '14 at 03:54

score 2 · Answer 2 · answered Sep 10 '14 at 00:11

Just for fun, here's a different way to do this immutably, by using a translation map.

If you wanted to replace everything that was in ltrFound, that would be easy:

tr = str.maketrans(ltrFound, '*' * len(ltrFound))
print(reveal.translate(tr))

But you want to do the opposite, replace everything that's not in ltrFound. And you don't want to build a translation table of all of the 100K+ characters that aren't s. So, what can you do?

You can build a table of the 6 characters that aren't in s but are in reveal:

notFound = ''.join(set(reveal) - set(ltrFound)) # 'adoprw'
tr = str.maketrans(notFound, '*' * len(notFound))
print(reveal.translate(tr))

The above is using Python 3.x; for 2.x, maketrans is a function in the string module rather than a classmethod of the str class (and there are a few other differences, but they don't matter here). So:

import string
notFound = ''.join(set(reveal) - set(ltrFound)) # 'adoprw'
tr = string.maketrans(notFound, '*' * len(notFound))
print(reveal.translate(tr))

hah I typed up most of this same answer and then put the re solution instead ! +1 — Joran Beasley, Sep 10 '14 at 02:12
@JoranBeasley: I wasn't going to write this until I realized that it's the obvious way to optimize his `replace`-based solution by doing all the `replace`s at once, and not searching for the distinct characters by iterating the string… and then, once I finished, I realized I had no idea how to explain that, and almost deleted it, but people had already upvoted it by then, so I figured I might as well let it stay. — abarnert, Sep 10 '14 at 03:53

score 1 · Answer 3 · answered Sep 09 '14 at 23:11

1

try this

re.sub("[^%s]"%guesses,"*",solution_string)

assuming guesses is a string

answered Sep 09 '14 at 23:11

Joran Beasley

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RegEx I can do... but I appreciate the suggestion nonetheless. – alcyone_pleiades Sep 09 '14 at 23:38

Replacing symbols using list comprehension

3 Answers3