Advanced array structure

Question

My dictionary has the following structure:

a = 'stringA'
b = 'stringB'
c = 'stringC'

endpoints = {
    'foo1': a + 'string1' + b,
    'foo2' : a + 'string2' + c + 'string3' + b,
}

My problem is the following: when I call endpoints['foo2'], I get the expected array value. However, when I change the value of, for instance, c between the array declaration and the usage of endpoints['foo2'], the value of c is not updated.

Any idea of why this happens and how can it be solved?

PS: I know this could be done creating a simple function, but I think that would be quite more inefficient.

@JeD There are no arrays here - or at all in python (unless you count numpy arrays and other non-native modules). endpoints is a dictionary. — jDo
– jDo, Commented Mar 28, 2016 at 9:51
@Alvaro Gomez "when I change the value of, for instance, c". Try posting the code you use to change the value of c — jDo
– jDo, Commented Mar 28, 2016 at 9:53
@KevinGuan yep Strings are immutable. You can get around it by defining a own class or putting the string into a list. — Jan
– Jan, Commented Mar 28, 2016 at 9:59
@JeD There no lists either :P But yeah, you're right. The values are fixed/immutable as soon as endpoints is created and it would take a re-definition of the particular entry to change it. @Alvaro would have to do c='newString' followed by endpoints['foo2'] = a + 'string2' + c + 'string3' + b or use a lambda function like Kevin Guan suggested. — jDo
– jDo, Commented Mar 28, 2016 at 10:06

Jan · Accepted Answer · 2016-03-28 10:25:15Z

2

You could do this:

a = ['stringA']
b = ['stringB']
c = ['stringC']

endpoints = {
    'foo1': a ,
    'foo2' : b
}

print(endpoints['foo1']) #returns "[stringA]"

a[0]='otherString'

print(endpoints['foo1']) #returns "[otherString]"

You can do this, because you can change the values in a list without changing the reference;

a and endpoints are still using the same space for a

This is not possible for pure Strings, because you can not change them without a new assignment. Strings in Python are immutable.

Edit: Another possibility would be to create your own string class.

This removes the [] brackets:

class MyStr:

    def __init__(self,val):
        self.val=val

    def __repr__(self):
    #this function is called by dict to get a string for the class
        return self.val

    def setVal(self,val):
        self.val=val

a=MyStr("abcd")
b={1:a}
print(b) #prints {1:"abcd"}
a.setVal("cdef")
print(b) #prints {1:"cdef"}

Disclaimer: As explained in the comments I am still using python 2.7

While Python 3 and 2.7 are mostly compatible, there might be some smaller bugs when trying to use this.

edited Mar 28, 2016 at 10:25

answered Mar 28, 2016 at 9:56

Jan

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9 Comments

Remi Guan Over a year ago

Aren't you getting TypeError: can only concatenate list (not "str") to list anyway? This is Python, not JavaScript or other programming languages like it.

Jan Over a year ago

@KevinGuan oh yeah, dammit

Ulrich Eckhardt Over a year ago

print is a function in nowadays Python.

Ulrich Eckhardt Over a year ago

It's not that the code is incorrect, @Kevin. It's just that it's written for Python 2.

jDo Over a year ago

@UlrichEckhardt But it'll usually work just fine in Python2 with the parentheses as well. You run the risk of inadvertently creating a tuple though. from __future__ import print_function if you want to be on the safe side.

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