What is the difference between
def delete_head1(t):
#t[:] = t[1:]
t = t[1:]
and
def delete_head2(t):
t[:] = t[1:]
#t = t[1:]
Why does the second one modify the input but not the first? That is, if I do
k=[1,2,3]
delete_head1(k)
print(k)
versus
k=[1,2,3]
delete_head2(k)
print(k)
The results are different.
As I understand, the first one, in its body, creates and refers to a local variable t. Why isn't the second one also referring to a local variable t?
If in the body of the function I have:
t = t[1:]
the interpreter recognizes the second t (the one after the equals sign) but not the first t (the one before the equals sign)?
But this is not the case if I write
t[:] = t[1:]
In this last statement it recognizes both t? What is the guiding principle here for Python?
Assignment statements behave differently depending on what the target it. You are correct that assignment to a name in a function create a local variable, but only one of your examples actually assigns to a name.
When the target is a simple name,
t = ...
this simply makes t refer to whatever object ... produces, and leaves the old value of t with one less reference.
When the target is a slice,
t[:] = ...
this is really syntactic sugar for
t = t.__setitem__(slice(None, None, None), ...)
t might refer to a new object afterwards, depending on what t.__setitem__ returns. But usually, this method returns the same object, and as a side effect alters the object in some way.
For lists, t[:] = some_list is basically equivalent to
t.clear()
t.extend(some_list)
t refers to the same list before and after, but the contents of that list change.
The full details of the syntax are in section 7.2, Assignment Statements of the language documentation. The semantics of list.__setitem__ in particular aren't spelled out in great detail, but can be inferred from the description of the syntax and the description of s[i:j:k] = t.
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.
Will
a, a = 2, 1
always result in a equal to 1? In other words, is tuple assignment guaranteed to be left-to-right?
The matter becomes relevant when we don't have just a, but a[i], a[j] and i and j may or may not be equal.
Yes, it is part of the python language reference that tuple assignment must take place left to right.
https://docs.python.org/2.3/ref/assignment.html
An assignment statement evaluates the expression list (remember that
this can be a single expression or a comma-separated list, the latter
yielding a tuple) and assigns the single resulting object to each of
the target lists, from left to right.
So all Python implementations should follow this rule (as confirmed by the experiments in the other answer).
Personally, I would still be hesitant to use this as it seems unclear to a future reader of the code.
How it works :
a, a = 2, 1
--> a does not exist, create variable a and set value to 2
--> a already exists, value of a changed to 1
When you have different variables, it works exactly the same way :
a, b, a = 1, 2, 3
--> a does not exist, create variable a and set value to 1
--> b does not exist, create variable b and set value to 2
--> a already exists, value of a changed to 3
I was reading the assignment statements in the Python docs ( http://docs.python.org/reference/simple_stmts.html#assignment-statements).
In that it is quoted that:
If the target is a target list enclosed in parentheses or in square brackets: The object must be an iterable with the same number of items as there are targets in the target list, and its items are assigned, from left to right, to the corresponding targets.
After reading it, I thought of writing a sample like this:
a = 5
b = 4
a, b = a + b, a
print a, b
My assumption was that a and b both should have the value of 9.
However, I am proven wrong. 'a' has the value of 9 and 'b' has the value of 5.
Can some one help me understand this better? Why the older value of 'a' is assigned rather than the new one? As per the docs, a's value will be assigned first right? Am I missing something?
All the expressions to the right of the assignment operator are evaluated before any of the assignments are made.
From the Python tutorial: First steps towards programming:
The first line contains a multiple assignment: the variables a and b simultaneously get the new values 0 and 1. On the last line this is used again, demonstrating that the expressions on the right-hand side are all evaluated first before any of the assignments take place. The right-hand side expressions are evaluated from the left to the right.
Emphasis mine.
Your code is functionally equivalent to the following:
a, b = 5 + 4, 5
print a, b
Python does not have a "comma operator" as in C. Instead, the comma indicates that a tuple should be constructed. The right-hand side of
a, b = a + b, a
is a tuple with th two items a + b and a.
On the left-hand side of an assignment, the comma indicates that sequence unpacking should be performed according to the rules you quoted: a will be assigned the first element of the tuple, b the second.
You can think of the assignments happening in parallel on copies rather than sequentially and in-place.
This is why in python you dont need a swap function:
a, b = b, a
works sufficiently without requiring a temp variable, c.
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.