I have a situation in Python(cough, homework) where I need to multiply EACH ELEMENT in a given list of objects a specified number of times and return the output of the elements. The problem is that the sample inputs given are of different types. For example, one case may input a list of strings whose elements I need to multiply while the others may be ints. So my return type needs to vary. I would like to do this without having to test what every type of object is. Is there a way to do this? I know in C# i could just use "var" but I don't know if such a thing exists in Python?
I realize that variables don't have to be declared, but in this case I can't see any way around it. Here's the function I made:
def multiplyItemsByFour(argsList):
output = ????
for arg in argsList:
output += arg * 4
return output
See how I need to add to the output variable. If I just try to take away the output assignment on the first line, I get an error that the variable was not defined. But if I assign it a 0 or a "" for an empty string, an exception could be thrown since you can't add 3 to a string or "a" to an integer, etc...
Here are some sample inputs and outputs:
Input: ('a','b') Output: 'aaaabbbb'
Input: (2,3,4) Output: 36
Thanks!
def fivetimes(anylist):
return anylist * 5
As you see, if you're given a list argument, there's no need for any assignment whatsoever in order to "multiply it a given number of times and return the output". You talk about a given list; how is it given to you, if not (the most natural way) as an argument to your function? Not that it matters much -- if it's a global variable, a property of the object that's your argument, and so forth, this still doesn't necessitate any assignment.
If you were "homeworkically" forbidden from using the * operator of lists, and just required to implement it yourself, this would require assignment, but no declaration:
def multiply_the_hard_way(inputlist, multiplier):
outputlist = []
for i in range(multiplier):
outputlist.extend(inputlist)
return outputlist
You can simply make the empty list "magicaly appear": there's no need to "declare" it as being anything whatsoever, it's an empty list and the Python compiler knows it as well as you or any reader of your code does. Binding it to the name outputlist doesn't require you to perform any special ritual either, just the binding (aka assignment) itself: names don't have types, only objects have types... that's Python!-)
Edit: OP now says output must not be a list, but rather int, float, or maybe string, and he is given no indication of what. I've asked for clarification -- multiplying a list ALWAYS returns a list, so clearly he must mean something different from what he originally said, that he had to multiply a list. Meanwhile, here's another attempt at mind-reading. Perhaps he must return a list where EACH ITEM of the input list is multiplied by the same factor (whether that item is an int, float, string, list, ...). Well then:
define multiply_each_item(somelist, multiplier):
return [item * multiplier for item in somelist]
Look ma, no hands^H^H^H^H^H assignment. (This is known as a "list comprehension", btw).
Or maybe (unlikely, but my mind-reading hat may be suffering interference from my tinfoil hat, will need to go to the mad hatter's shop to have them tuned) he needs to (say) multiply each list item as if they were the same type as the first item, but return them as their original type, so that for example
>>> mystic(['zap', 1, 23, 'goo'], 2)
['zapzap', 11, 2323, 'googoo']
>>> mystic([23, '12', 15, 2.5], 2)
[46, '24', 30, 4.0]
Even this highly-mystical spec COULD be accomodated...:
>>> def mystic(alist, mul):
... multyp = type(alist[0])
... return [type(x)(mul*multyp(x)) for x in alist]
...
...though I very much doubt it's the spec actually encoded in the mysterious runes of that homework assignment. Just about ANY precise spec can be either implemented or proven to be likely impossible as stated (by requiring you to solve the Halting Problem or demanding that P==NP, say;-). That may take some work ("prove the 4-color theorem", for example;-)... but still less than it takes to magically divine what the actual spec IS, from a collection of mutually contradictory observations, no examples, etc. Though in our daily work as software developer (ah for the good old times when all we had to face was homework!-) we DO meet a lot of such cases of course (and have to solve them to earn our daily bread;-).
EditEdit: finally seeing a precise spec I point out I already implemented that one, anyway, here it goes again:
def multiplyItemsByFour(argsList):
return [item * 4 for item in argsList]
EditEditEdit: finally/finally seeing a MORE precise spec, with (luxury!-) examples:
Input: ('a','b') Output: 'aaaabbbb' Input: (2,3,4) Output: 36
So then what's wanted it the summation (and you can't use sum as it wouldn't work on strings) of the items in the input list, each multiplied by four. My preferred solution:
def theFinalAndTrulyRealProblemAsPosed(argsList):
items = iter(argsList)
output = next(items, []) * 4
for item in items:
output += item * 4
return output
If you're forbidden from using some of these constructs, such as built-ins items and iter, there are many other possibilities (slightly inferior ones) such as:
def theFinalAndTrulyRealProblemAsPosed(argsList):
if not argsList: return None
output = argsList[0] * 4
for item in argsList[1:]:
output += item * 4
return output
For an empty argsList, the first version returns [], the second one returns None -- not sure what you're supposed to do in that corner case anyway.
Very easy in Python. You need to get the type of the data in your list - use the type() function on the first item - type(argsList[0]). Then to initialize output (where you now have ????) you need the 'zero' or nul value for that type. So just as int() or float() or str() returns the zero or nul for their type so to will type(argsList[0])() return the zero or nul value for whatever type you have in your list.
So, here is your function with one minor modification:
def multiplyItemsByFour(argsList):
output = type(argsList[0])()
for arg in argsList:
output += arg * 4
return output
Works with::
argsList = [1, 2, 3, 4] or [1.0, 2.0, 3.0, 4.0] or "abcdef" ... etc,
Are you sure this is for Python beginners? To me, the cleanest way to do this is with reduce() and lambda, both of which are not typical beginner tools, and sometimes discouraged even for experienced Python programmers:
def multiplyItemsByFour(argsList):
if not argsList:
return None
newItems = [item * 4 for item in argsList]
return reduce(lambda x, y: x + y, newItems)
Like Alex Martelli, I've thrown in a quick test for an empty list at the beginning which returns None. Note that if you are using Python 3, you must import functools to use reduce().
Essentially, the reduce(lambda...) solution is very similar to the other suggestions to set up an accumulator using the first input item, and then processing the rest of the input items; but is simply more concise.
My guess is that the purpose of your homework is to expose you to "duck typing". The basic idea is that you don't worry about the types too much, you just worry about whether the behaviors work correctly. A classic example:
def add_two(a, b):
return a + b
print add_two(1, 2) # prints 3
print add_two("foo", "bar") # prints "foobar"
print add_two([0, 1, 2], [3, 4, 5]) # prints [0, 1, 2, 3, 4, 5]
Notice that when you def a function in Python, you don't declare a return type anywhere. It is perfectly okay for the same function to return different types based on its arguments. It's considered a virtue, even; consider that in Python we only need one definition of add_two() and we can add integers, add floats, concatenate strings, and join lists with it. Statically typed languages would require multiple implementations, unless they had an escape such as variant, but Python is dynamically typed. (Python is strongly typed, but dynamically typed. Some will tell you Python is weakly typed, but it isn't. In a weakly typed language such as JavaScript, the expression 1 + "1" will give you a result of 2; in Python this expression just raises a TypeError exception.)
It is considered very poor style to try to test the arguments to figure out their types, and then do things based on the types. If you need to make your code robust, you can always use a try block:
def safe_add_two(a, b):
try:
return a + b
except TypeError:
return None
See also the Wikipedia page on duck typing.
Python is dynamically typed, you don't need to declare the type of a variable, because a variable doesn't have a type, only values do. (Any variable can store any value, a value never changes its type during its lifetime.)
def do_something(x):
return x * 5
This will work for any x you pass to it, the actual result depending on what type the value in x has. If x contains a number it will just do regular multiplication, if it contains a string the string will be repeated five times in a row, for lists and such it will repeat the list five times, and so on. For custom types (classes) it depends on whether the class has an operation defined for the multiplication operator.
You don't need to declare variable types in python; a variable has the type of whatever's assigned to it.
EDIT:
To solve the re-stated problem, try this:
def multiplyItemsByFour(argsList):
output = argsList.pop(0) * 4
for arg in argsList:
output += arg * 4
return output
(This is probably not the most pythonic way of doing this, but it should at least start off your output variable as the right type, assuming the whole list is of the same type)
You gave these sample inputs and outputs:
Input: ('a','b') Output: 'aaaabbbb' Input: (2,3,4) Output: 36
I don't want to write the solution to your homework for you, but I do want to steer you in the correct direction. But I'm still not sure I understand what your problem is, because the problem as I understand it seems a bit difficult for an intro to Python class.
The most straightforward way to solve this requires that the arguments be passed in a list. Then, you can look at the first item in the list, and work from that. Here is a function that requires the caller to pass in a list of two items:
def handle_list_of_len_2(lst):
return lst[0] * 4 + lst[1] * 4
Now, how can we make this extend past two items? Well, in your sample code you weren't sure what to assign to your variable output. How about assigning lst[0]? Then it always has the correct type. Then you could loop over all the other elements in lst and accumulate to your output variable using += as you wrote. If you don't know how to loop over a list of items but skip the first thing in the list, Google search for "python list slice".
Now, how can we make this not require the user to pack up everything into a list, but just call the function? What we really want is some way to accept whatever arguments the user wants to pass to the function, and make a list out of them. Perhaps there is special syntax for declaring a function where you tell Python you just want the arguments bundled up into a list. You might check a good tutorial and see what it says about how to define a function.
Now that we have covered (very generally) how to accumulate an answer using +=, let's consider other ways to accumulate an answer. If you know how to use a list comprehension, you could use one of those to return a new list based on the argument list, with the multiply performed on each argument; you could then somehow reduce the list down to a single item and return it. Python 2.3 and newer have a built-in function called sum() and you might want to read up on that. [EDIT: Oh drat, sum() only works on numbers. See note added at end.]
I hope this helps. If you are still very confused, I suggest you contact your teacher and ask for clarification. Good luck.
P.S. Python 2.x have a built-in function called reduce() and it is possible to implement sum() using reduce(). However, the creator of Python thinks it is better to just use sum() and in fact he removed reduce() from Python 3.0 (well, he moved it into a module called functools).
P.P.S. If you get the list comprehension working, here's one more thing to think about. If you use a list comprehension and then pass the result to sum(), you build a list to be used once and then discarded. Wouldn't it be neat if we could get the result, but instead of building the whole list and then discarding it we could just have the sum() function consume the list items as fast as they are generated? You might want to read this: Generator Expressions vs. List Comprehension
EDIT: Oh drat, I assumed that Python's sum() builtin would use duck typing. Actually it is documented to work on numbers, only. I'm disappointed! I'll have to search and see if there were any discussions about that, and see why they did it the way they did; they probably had good reasons. Meanwhile, you might as well use your += solution. Sorry about that.
EDIT: Okay, reading through other answers, I now notice two ways suggested for peeling off the first element in the list.
For simplicity, because you seem like a Python beginner, I suggested simply using output = lst[0] and then using list slicing to skip past the first item in the list. However, Wooble in his answer suggested using output = lst.pop(0) which is a very clean solution: it gets the zeroth thing on the list, and then you can just loop over the list and you automatically skip the zeroth thing. However, this "mutates" the list! It's better if a function like this does not have "side effects" such as modifying the list passed to it. (Unless the list is a special list made just for that function call, such as a *args list.) Another way would be to use the "list slice" trick to make a copy of the list that has the first item removed. Alex Martelli provided an example of how to make an "iterator" using a Python feature called iter(), and then using iterator to get the "next" thing. Since the iterator hasn't been used yet, the next thing is the zeroth thing in the list. That's not really a beginner solution but it is the most elegant way to do this in Python; you could pass a really huge list to the function, and Alex Martelli's solution will neither mutate the list nor waste memory by making a copy of the list.
No need to test the objects, just multiply away!
'this is a string' * 6
14 * 6
[1,2,3] * 6
all just work
Try this:
def timesfourlist(list):
nextstep = map(times_four, list)
sum(nextstep)
map performs the function passed in on each element of the list(returning a new list) and then sum does the += on the list.
If you just want to fill in the blank in your code, you could try setting object=arglist[0].__class__() to give it the zero equivalent value of that class.
>>> def multiplyItemsByFour(argsList):
output = argsList[0].__class__()
for arg in argsList:
output += arg * 4
return output
>>> multiplyItemsByFour('ab')
'aaaabbbb'
>>> multiplyItemsByFour((2,3,4))
36
>>> multiplyItemsByFour((2.0,3.3))
21.199999999999999
This will crash if the list is empty, but you can check for that case at the beginning of the function and return whatever you feel appropriate.
Thanks to Alex Martelli, you have the best possible solution:
def theFinalAndTrulyRealProblemAsPosed(argsList):
items = iter(argsList)
output = next(items, []) * 4
for item in items:
output += item * 4
return output
This is beautiful and elegant. First we create an iterator with iter(), then we use next() to get the first object in the list. Then we accumulate as we iterate through the rest of the list, and we are done. We never need to know the type of the objects in argsList, and indeed they can be of different types as long as all the types can have operator + applied with them. This is duck typing.
For a moment there last night I was confused and thought that you wanted a function that, instead of taking an explicit list, just took one or more arguments.
def four_x_args(*args):
return theFinalAndTrulyRealProblemAsPosed(args)
The *args argument to the function tells Python to gather up all arguments to this function and make a tuple out of them; then the tuple is bound to the name args. You can easily make a list out of it, and then you could use the .pop(0) method to get the first item from the list. This costs the memory and time to build the list, which is why the iter() solution is so elegant.
def four_x_args(*args):
argsList = list(args) # convert from tuple to list
output = argsList.pop(0) * 4
for arg in argsList:
output += arg * 4
return output
This is just Wooble's solution, rewritten to use *args.
Examples of calling it:
print four_x_args(1) # prints 4
print four_x_args(1, 2) # prints 12
print four_x_args('a') # prints 'aaaa'
print four_x_args('ab', 'c') # prints 'ababababcccc'
Finally, I'm going to be malicious and complain about the solution you accepted. That solution depends on the object's base class having a sensible null or zero, but not all classes have this. int() returns 0, and str() returns '' (null string), so they work. But how about this:
class NaturalNumber(int):
"""
Exactly like an int, but only values >= 1 are possible.
"""
def __new__(cls, initial_value=1):
try:
n = int(initial_value)
if n < 1:
raise ValueError
except ValueError:
raise ValueError, "NaturalNumber() initial value must be an int() >= 1"
return super(NaturalNumber, cls).__new__ (cls, n)
argList = [NaturalNumber(n) for n in xrange(1, 4)]
print theFinalAndTrulyRealProblemAsPosed(argList) # prints correct answer: 24
print NaturalNumber() # prints 1
print type(argList[0])() # prints 1, same as previous line
print multiplyItemsByFour(argList) # prints 25!
Good luck in your studies, and I hope you enjoy Python as much as I do.
Related
Hope you're all well with the caotic world we're living...
This might be a very beginner level question, but I'd like to understand why It is like that.
Let's say I have a list of complex:
myList = [(1.231 +2.254j), (2.875 +23.543j), ...]
I've been trying to round the values with this function:
def round_complex(x, digits):
return complex(round(x.real, digits), round(x.imag, digits))
And for doing so, I've tried this:
for item in myList:
item = round_complex(item, 2)
Expecting that myList values get changed, for example:
myList = [(1.23 +2.25j), (2.88 +23.54j), ...]
But, It does not work.
I've also tried with a more simple example, like a list of floats and the base round function from python. It also does not work.
Is there a way for me to change a value of an iterable object with this kind of for loop (for-in)?
Or do I really have to do this:
for i in range(len(myList)):
myList[i] = round_complex(myList[i], 2)
The simple answer is: NO.
Python uses a mechanism, which is known as "Call-by-Object", sometimes also called "Call by Object Reference" or "Call by Sharing" when pass function parameters.
If you pass immutable arguments like integers, strings or tuples to a function, the passing acts like call-by-value. The object reference is passed to the function parameters. They can't be changed within the function, because they can't be changed at all, i.e. they are immutable. It's different, if we pass mutable arguments. They are also passed by object reference, but they can be changed in place within the function.
So, after your iterate the list, the value (1.231 +2.254j) would be a immutable argument which your change won't affect the outside variable. But if you pass the value like [1.231 +2.254j] to function, then it will make effect like next:
test.py:
myList2 = [[(1.231 +2.254j)], [(2.875 +23.543j)]]
print(myList2)
def round_complex(x, digits):
return complex(round(x.real, digits), round(x.imag, digits))
for item2 in myList2:
item2[0] = round_complex(item2[0], 2)
print(myList2)
Execution:
$ python3 test.py
[[(1.231+2.254j)], [(2.875+23.543j)]]
[[(1.23+2.25j)], [(2.88+23.54j)]]
In a word, for you scenario, if you insist organize your input data as that & iterate with that way, you can't change the outside value directly inside the function.
You may refers to this to learn more.
The thing I understand by reading for question that you want to Assign the values of i in myList. For doing so you can use append i.e
for i in mylist:
mylist.append(round_complex(i, 2))
I'm new to Python and I'm messing around with this and I don't really know why when I change the brackets to parenthesis I get an error, or why I can't just use len(text - 1).
I'm looking at this code:
def reverse(text):
result = ""
length = len(text)
for i in text:
result += text[length - 1]
length -= 1
return result
Any help with understanding this is greatly appreciated!
Thanks!
You can't use text - 1 because that doesn't make sense; you cannot take 1 from a string.
Now, as for len(...) this is a function so has () for the call rather than []. You pass it the thing you want the length of.
text[length - 1] is indexing into the string, at a position, and follows the list syntax for indexing or sub-scripting.
When you use len(text - 1) you try to subtract int (1) from str (text). It is insupported operation in python (and moreover, impossible). To get a part of string you need you must use text[length - 1].
Python Parentheses
Parentheses play many different roles in Python these are some of the main roles they play:
The mathematical role:
Python parentheses act like parentheses in math as they are at the top of the Python Priority Precedence
This means that this:
>>> 3 + 4 * 2
returns:
12
Whereas with parentheses:
>>> (3 + 4) * 2
returns:
14
But that's not all, their priority also expands to Boolean expressions:
for example:
False and False or True and True
evaluates to True as and is executed before or. However, if you add some parentheses:
False and (False or True) and True
It evaluates to False as the or is executed before the and
Tuple
In python, when you put something in a tuple you use () notation.
Functions
When you declare or call a function you always need to add the parentheses after the function name. Think of them as a basket to put the arguments in. If you forget them the Python interpreter will think that you are calling a variable for example:
list
This is a variable called list and does nothing special
list() #Empty basket
This, however, is a call to a function as there is a "basket"
Square Brackets
Square Brackets also have quite a few roles:
Lists
In python, you use square brackets if you want to declare a list instead of a tuple.
List comprehension
List comprehension is actually pretty complicated so read this for more information but just know that it uses square brackets
Looking up
The main use of square brackets is to look up a value inside a list, tuple, or dictionary. Think of it like the google search bar: you write what you want and it tells you what it has. For example:
tuple = (2, 4)
if you want to get 4 you will need to look up the 2nd value of the tuple:
tuple[1] #The first value is 0
Slicing
Slicing is simply the idea of taking only certain parts of a list (or tuple, dictionary or even string). Here is an explanation by Greg Hewgill (https://stackoverflow.com/a/509295/7541446):
There is also the step value, which can be used with any of the above:
a[start:end:step] # start through not past end, by step
The key point to remember is that the :end value represents the first
value that is not in the selected slice. So, the difference beween end
and start is the number of elements selected (if step is 1, the
default).
The other feature is that start or end may be a negative number, which
means it counts from the end of the array instead of the beginning.
So:
a[-1] # last item in the array a[-2:] # last two items in the
array a[:-2] # everything except the last two items
Python is kind to the programmer if there are fewer items than you ask
for. For example, if you ask for a[:-2] and a only contains one
element, you get an empty list instead of an error. Sometimes you
would prefer the error, so you have to be aware that this may happen.
I hope this provided useful insight to explaining the difference between parentheses and square brackets.
This means that in your question len() is a function where you are putting text inside the basket. However, when you call text[length-1] you are looking up the value at position length-1
The python builtin function len() returns the length in numbers of the object in argument e.g
temp = [1, 2, 3, 4]
length = len(temp)
then the len() will return 4 in this case. but if someone write
length = len(temp-1)
then
temp-1
is not a valid object, therefor you cannot use it.
The reason you can't do len(text-1) is because text is a string type you are trying to reverse, and being a string you cannot combine it with numbers(unless they are a string, but thats a different story) without getting an error. Therefore you do len(text)-1 because len(text) would equal the length of whatever the text is(lets say four), and then you can subtract 1 from 4 because 4 is an integer.
The reason you need brackets and not parentheses when you are doing text[length-1] is because in python trying to get a single value out of a string requires the use of string[] and putting a position in the string inside the []. You use partakes to call functions like print(string) or len(text), so putting text(length-1) would create an error that basically says the program doesn't have a function named "text" and it doesn't know what to do.
Hope this helps. Tell me if you have any more questions.
I'm trying to write a short program which allows the user to input a list of numbers into an input() function, and then using the add_25 function add 25 to each item in a list.
I get the following error when the program runs: TypeError: 'tuple' object does not support item assignment
I tried dividing the numbers using a comma. This is the program:
testlist = [2,6,2]
def add_25(mylist):
for i in range(0, len(mylist)):
mylist[i] = mylist[i] + 25
return mylist
print add_25(testlist)
actual_list = input("Please input a series of numbers, divided by a comma:")
print add_25(actual_list)
In Python 2 input() will eval the string and in this case it will create a tuple, and as tuples are immutable you'll get that error.
>>> eval('1, 2, 3')
(1, 2, 3)
It is safer to use raw_input with a list-comprehension here:
inp = raw_input("Please input a series of numbers, divided by a comma:")
actual_list = [int(x) for x in inp.split(',')]
Or if you're not worried about user's input then simply convert the tuple to list by passing it to list().
Also note that as you're trying to update the list in-place inside of the function it makes no sense to return the list unless you want to assign another variable to the same list object. Either return a new list or don't return anything.
The function input reads a string and evaluates it as a Python expression. Thus, the comma-separated list becomes a tuple of values, these are passed to add_25(), and this function tries to assign to mylist[i] something.
And tuples are immutable, they do not support item assignment (on purpose).
You could use actual_list = list(input(...)) to convert the tuple to a list (which supports item assignment).
But every time someone uses input(), one has to warn him: input() is a security risk. It evaluates the input of the user and thus might execute arbitrary things the user typed. This means that your program will perform what the user asks it to, with your permissions. This is normally not what is considered a good design.
If you always will be the only user or if you trust all users of your program completely, then so be it.
Besides all the input() aspects covered in the other answers, I'd like to add this completely different aspect:
Your function add_25() is probably not supposed to change its input. Yours does, or tries to, and fails because tuples do not allow that.
But you actually do not have to change the input (and you should not, because this is not good style due to its ugly side-effects). Instead you could just return a new tuple:
def add_25(mytuple):
return tuple(x + 25 for x in mytuple)
This way, nothing is assigned to a tuple, just a new tuple is created and returned.
def test_tuples_of_one_look_peculiar(self):
self.assertEqual( __,(1).__class__)
self.assertEqual(__, (1,).__class__)
I was wondering if someone can help me to fix the error my code for quick sort has:
It does not compile and highlights the last line of the code in red.
I can not figure out what is wrong. sort is already defined as a function so why is it highlighted as red?
def sort(*myarray):
less = []
equal = []
greater = []
if len(myarray) > 1:
pivot = myarray[0]
for x in myarray:
if x < pivot:
less.append(x)
if x == pivot:
equal.append(x)
if x > pivot:
greater.append(x)
return sort(less)+sort(equal)+sort(greater)
else:
return myarray
print sort([12,4,5,6,7,3,1,15])
You're defining the function as taking a variable number of arguments (the *myarray bit), but then using myarray inside as a single argument (the list to sort), when it is a list containing the list to sort.
You probably should remove the * from your function parameter. This questions esplains it quite thoroughly.
You could keep the *, but then you would have to play a bit with tuple unpacking to get the same result.
edit
Although the above is true, this might not be the issue you're encountering.
IDLE will give you the invalid syntax error on the ast line, because in interactive mode - with lines starting with >>>, it accepts only one statement at a time. In your case that statement is the sort() definition.
Try hitting enter 2 times after the function definition, this should get you back to the repl, where you can introduce another statement (print sort([12,4,5,6,7,3,1,15]))
There are a couple things wrong which makes me curious how you are testing this:
Python code is not "compiled", it is interpreted. (Okay, not precisely true; it's parsed into a sort of byte-code; still, it's not compiled in the same sense as a language such as C, where the entire program has to be converted into machine instructions before any of it can be run.) Also you mention the last line of code is highlighted in red -- by what?
This code actually works, but only if you remote the star/asterisk in front of myarray in def sort(*myarray):. Otherwise it actually returns a single-element tuple containing the original array.
Assuming you have two or more elements that equal a pivot at some point, you get an infinite loop, because you will get: equal = [x,x] (two elements at least), and then invoke sort([x,x]), which in its turn will take x as a pivot, and create equal = [x,x], and cause sort([x,x]), ....
Simple solution to this problem: What should be the output of the sort(equal)? How do you sort a list of identical elements?
Edit: Well, your comments show that you are looking for a different problem, but I'll leave it here because it explains a different issue you have with your code and should be solved.
If it is a function for quick sorting, can you really use the function sort in it?
Wouldn't something like this work?
def qsort(list):
pivind=0
left, right, pivot= [], [], []
for x in list:
if list[pivind]==x: pivot.append(x)
elif list[pivind]>x: left.append(x)
else: right.append(x)
if len(left)>1: left=qsort(left)
if len(right)>1: right=qsort(right)
return (left + pivot + right)
Let's say I have a function f() that takes a list and returns a mutation of that list. If I want to apply that function to five member variables in my class instance (i), I can do this:
for x in [i.a, i.b, i.c, i.d, i.e]:
x[:] = f(x)
1) Is there a more elegant way? I don't want f() to modify the passed list.
2) If my variables hold a simple integer (which won't work with the slice notation), is there also a way? (f() would also take & return an integer in this case)
Another solution, though it's probably not elegant:
for x in ['a', 'b', 'c', 'd', 'e']:
setattr(i, x, f(getattr(i, x)))
Python doesn't have pass by reference. The best you can do is write a function which constructs a new list and assign the result of the function to the original list. Example:
def double_list(x):
return [item * 2 for item in x]
nums = [1, 2, 3, 4]
nums = double_list(nums) # now [2, 4, 6, 8]
Or better yet:
nums = map(lambda x: x * 2, nums)
Super simple example, but you get the idea. If you want to change a list from a function you'll have to return the altered list and assign that to the original.
You might be able to hack up a solution, but it's best just to do it the normal way.
EDIT
It occurs to me that I don't actually know what you're trying to do, specifically. Perhaps if you were to clarify your specific task we could come up with a solution that Python will permit?
Ultimately, what you want to do is incompatible with the way that Python is structured. You have the most elegant way to do it already in the case that your variables are lists but this is not possible with numbers.
This is because variables do not exist in Python. References do. So i.x is not a list, it is a reference to a list. Likewise, if it references a number. So if i.x references y, then i.x = z doesn't actually change the value y, it changes the location in memory that i.x points to.
Most of the time, variables are viewed as boxes that hold a value. The name is on the box. In python, values are fundamental and "variables" are just tags that get hung on a particular value. It's very nice once you get used to it.
In the case of a list, you can use use slice assignment, as you are already doing. This will allow all references to the list to see the changes because you are changing the list object itself. In the case of a number, there is no way to do that because numbers are immutable objects in Python. This makes sense. Five is five and there's not much that you can do to change it. If you know or can determine the name of the attribute, then you can use setattr to modify it but this will not change other references that might already exist.
As Rafe Kettler says, if you can be more specific about what you actually want to do, then we can come up with a simple elegant way to do it.