I have a list a = [1,2,3,4,5]
I don't understand why the following code doesn't produce [2,3,4,5,1]
a[1:].append(a[0])
I tried reading up on the append() method as well as list slicing in Python, but found no satisfactory response.
a[1:] gives you a whole new list, but you're not assigning it to any variable so it it just thrown away after that line. You should assign it to something (say, b) and then append to it (otherwise append would change the list but return nothing):
a = [1,2,3,4,5]
b = a[1:]
b.append(a[0])
And now b is your desired output [2,3,4,5,1]
I think here you just not really understand append function. Just like my answer in Passing a variable from one file into another as a class variable after inline modification, append is an in-place operation which just update the original list and return None.
Your chain call a[1:].append(a[0]) will return the last call return value in the chain, so return append function value, which is None.
Just like #flakes comment in another answer, a[1:] + a[:1] will return your target value. Also you can try a[1:][1:][1:] which will return some
result. So the key point is the append function is in-place function.
See python more on list
You might have noticed that methods like insert, remove or sort that only modify the list have no return value printed – they return the default None. 1 This is a design principle for all mutable data structures in Python.
First, append the first element a[0] to the list
a.append(a[0])
and then exclude the first element
a[1:]
I am a complete newbie to python and attempting to pass an array as an argument to a python function that declares a list/array as the parameter.
I am sure I am declaring it wrong,
here goes:
def dosomething(listparam):
#do something here
dosomething(listargument)
Clearly this is not working, what am I doing wrong?
Thanks
What you have is on the right track.
def dosomething( thelist ):
for element in thelist:
print element
dosomething( ['1','2','3'] )
alist = ['red','green','blue']
dosomething( alist )
Produces the output:
1
2
3
red
green
blue
A couple of things to note given your comment above: unlike in C-family languages, you often don't need to bother with tracking the index while iterating over a list, unless the index itself is important. If you really do need the index, though, you can use enumerate(list) to get index,element pairs, rather than doing the x in range(len(thelist)) dance.
Maybe you want to unpack elements of an array, I don't know if I got it, but below an example:
def my_func(*args):
for a in args:
print(a)
my_func(*[1,2,3,4])
my_list = ['a','b','c']
my_func(*my_list)
I guess I'm unclear about what the OP was really asking for... Do you want to pass the whole array/list and operate on it inside the function? Or do you want the same thing done on every value/item in the array/list. If the latter is what you wish I have found a method which works well.
I'm more familiar with programming languages such as Fortran and C, in which you can define elemental functions which operate on each element inside an array. I finally tracked down the python equivalent to this and thought I would repost the solution here. The key is to 'vectorize' the function. Here is an example:
def myfunc(a,b):
if (a>b): return a
else: return b
vecfunc = np.vectorize(myfunc)
result=vecfunc([[1,2,3],[5,6,9]],[7,4,5])
print(result)
Output:
[[7 4 5]
[7 6 9]]
This question already has answers here:
Why do these list operations (methods: clear / extend / reverse / append / sort / remove) return None, rather than the resulting list?
(6 answers)
Closed 5 months ago.
I'm having an issue considering the built-in Python List-methods.
As I learned Python, I always thought Python mutators, as any value class mutators should do, returned the new variable it created.
Take this example:
a = range(5)
# will give [0, 1, 2, 3, 4]
b = a.remove(1)
# as I learned it, b should now be [0, 2, 3, 4]
# what actually happens:
# a = [0, 2, 3, 4]
# b = None
The main problem with this list mutator not returning a new list, is that you cannot to multiple mutations subsequently.
Say I want a list ranging from 0 to 5, without the 2 and the 3.
Mutators returning new variables should be able to do it like this:
a = range(5).remove(2).remove(3)
This sadly isn't possible, as range(5).remove(2) = None.
Now, is there a way to actually do multiple mutations on lists like I wanna do in my example? I think even PHP allows these types of subsequent mutations with Strings.
I also can't find a good reference on all the built-in Python functions. If anyone can find the actual definition (with return values) of all the list mutator methods, please let me know. All I can find is this page: http://docs.python.org/tutorial/datastructures.html
Rather than both mutating and returning objects, the Python library chooses to have just one way of using the result of a mutator. From import this:
There should be one-- and preferably only one --obvious way to do it.
Having said that, the more usual Python style for what you want to do is using list comprehensions or generator expressions:
[x for x in range(5) if x != 2 and x != 3]
You can also chain these together:
>>> [x for x in (x for x in range(5) if x != 2) if x != 3]
[0, 1, 4]
The above generator expression has the added advantage that it runs in O(n) time because Python only iterates over the range() once. For large generator expressions, and even for infinite generator expressions, this is advantageous.
Many methods of list and other mutable types intentionally return None so that there is no question in your mind as to whether you are creating a new object or mutating an existing object. The only thing that could be happening is mutation since, if a new object were created, it would have to be returned by the method, and it is not returned.
As you may have noticed, the methods of str that edit the string do return the new string, because strings are immutable and a new string is always returned.
There is of course nothing at all keeping you from writing a list subclass that has the desired behavior on .append() et al, although this seems like rather a heavy hammer to swing merely to allow you to chain method calls. Also, most Python programmers won't expect that behavior, making your code less clear.
In Python, essentially all methods that mutate the object return None.
That's so you don't accidentally think you've got a new object as a result.
You mention
I think even PHP allows these types of subsequent mutations with Strings.
While I don't remember about PHP, with string manipulations, you can chain method calls, because strings are immutable, so all string methods return a new string, since they don't (can't) change the one you call them on.
>>> "a{0}b".format(3).center(5).capitalize().join('ABC').swapcase()
'a A3B b A3B c'
Neither Python or php have built-in "mutators". You can simply write multiple lines, like
a = list(range(5))
a.remove(2)
a.remove(3)
If you want to generate a new list, copy the old one beforehand:
a = list(range(5))
b = a[:]
b.remove(2)
Note that in most cases, a singular call to remove indicates you should have used a set in the first place.
To remove multiple mutations on lists as you want to do on your example, you can do :
a = range(5)
a = [i for j, i in enumerate(a) if j not in [2, 3]]
a will be [0, 1, 4]
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.
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.