11

I have a a user-inputted polynomial and I only want to use it if it only has characters in the string 1234567890^-+x.

How can I check if it does or not without using external packages? I only want to use built-in Python 2.5 functions.

I am writing a program that runs on any Mac without needing external packages.

senshin
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user2658538
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6 Answers6

18

Here are some odd ;-) ways to do it:

good = set('1234567890^-+x')

if set(input_string) <= good:
    # it's good
else:
    # it's bad

or

if input_string.strip('1234567890^-+x'):
    # it's bad!
else:
    # it's good
Tim Peters
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    I love the set subset approach :-) Care to take a `timeit` potshot at it? – Martijn Pieters Dec 22 '13 at 03:51
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    @MartijnPieters, not a chance - there are sooooo many ways to do this I don't want to spend half an hour organizing them all ;-) – Tim Peters Dec 22 '13 at 03:53
  • @user2357112, just because I can never remember what `.issuperset()` does, exactly, without looking it up :-( But I understand `<=` at once, and as Martijn says, they're the same thing in the end. `.issuperset()` is likely slower due to the method lookup expense. – Tim Peters Dec 22 '13 at 03:54
  • `issuperset()` is simply the inverse of `issubset()`; the `>=` to your `<=`. – Martijn Pieters Dec 22 '13 at 03:55
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    @MartijnPieters, sure, but with the method spellings I can never remember **which one** is `<=` and which `>=` without looking it up. So I avoid them. `<=` and `>=` are obvious to me. – Tim Peters Dec 22 '13 at 03:56
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    @MartijnPieters: `issuperset` would... wait, apparently it doesn't short-circuit. I thought it would do the test without building a set from the `input_string`, but when I tried `{0}.issuperset(xrange(1000000000))` a few seconds ago, it started eating memory in a way that seems to indicate it turns the input into a set. I guess there's no performance advantage. – user2357112 Dec 22 '13 at 03:57
10

Use a regular expression:

import re

if re.match('^[-0-9^+x]*$', text):
    # Valid input

The re module comes with Python 2.5, and is your fastest option.

Demo:

>>> re.match('^[-0-9^+x]*$', '1x2^4-2')
<_sre.SRE_Match object at 0x10f0b6780>
Martijn Pieters
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4
  1. You can convert the valid chars to a set, as sets offer faster lookup

  2. Then you can use all function like this

    valid_chars = set("1234567890^-+x")  # Converting to a set
if all(char in valid_chars for char in input_string):
    # Do stuff if input is valid

  3. We can convert the input string also a set and check if all characters in the inputstring is in the valid list.

    valid_chars = set("1234567890^-+x")  # Converting to a set
if set(input_string).issubset(valid_chars):
    # Do stuff if input is valid
thefourtheye
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4

What about just convert both the string into set and checking input_set is subset of good_set as below:

>>> good_set = set('1234567890^-+x')
>>> input_set1 = set('xajfb123')
>>> input_set2 = set('122-32+x')
>>> input_set1.issubset(good_set)
False
>>> input_set2.issubset(good_set)
True
>>>
James Sapam
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1

Yet another way to do it, now using string.translate():

>>> import string
>>> all_chars = string.maketrans('', '')
>>> has_only = lambda s, valid_chars: not s.translate(all_chars, valid_chars)
>>> has_only("abc", "1234567890^-+x.")
False
>>> has_only("x^2", "1234567890^-+x.")
True

It is not the most readable way. It should be one of the fastest if you need it.

jfs
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0
whitelist = '1234567890^-+x'

str = 'x^2+2x+1'
min([ch in whitelist for ch in str])
True


str='x**2 + 1' 
min([ch in whitelist for ch in str])
False
James King
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