I am confused about when using a while-loop or for-loop is better? I am particularly worried about producing optimized answers to coding questions. I find myself solving problems just to find out a while-loop would've been faster but am confused about what leads people to choose to use it instead of a for-loop, like what criteria should I be looking for?
Here's an example of a coding question I answered which checks if a string of parentheses is balanced.
def parenCheck(str):
stack = Stack()
for x in str:
if x == '(':
stack.push(x)
else:
if stack.isEmpty():
return False
else:
stack.pop()
return stack.isEmpty()
Here is the answer I saw for it , which I know is faster because it doesnt use a for-loop:
def parChecker(symbolString):
s = Stack()
balanced = True
index = 0
while index < len(symbolString) and balanced:
symbol = symbolString[index]
if symbol == "(":
s.push(symbol)
else:
if s.isEmpty():
balanced = False
else:
s.pop()
index = index + 1
if balanced and s.isEmpty():
return True
else:
return False
For the same number of iterations, and content, my guess is that a for loop and wihile have practically the same speed. But I haven't tested that - mostly I answer numpy questions where we try to avoid either loop (preferring iterations in compiled code).
Your examples show the basic case where for results in cleaner, if not faster, code:
for x in an_iterator:
<do something>
versus
i = 0
while i < len(an_iterator):
x = an_iterator[x]
<do something>
i += 1
If you must have an index as well, you can use:
for i, x in enumerate(an_iterator):
....
That ability to iterate directly on things like lists and dictionaries was the main neat feature that caught my attention when I first saw Python in the 1990.
A common subcase of a for loop accumulates values, so common, that Python provides the popular list comprehension, and extended the syntax to generator expressions and dictionary comprehensions.
while still has its uses. Other languages have do while and do until variants, which attempt to streamline stepping the variable and testing. Python has just the one version with separate steps. The new walrus operator has the potential of cleaning that up:
https://docs.python.org/3/whatsnew/3.8.html#assignment-expressions
while (block := f.read(256)) != '':
process(block)
while is most useful when the steps aren't regular, or from a well defined iterator. for can break and continue, but otherwise the sequence is fixed (at the start of iteration).
In addition to enumerate, zip lets us iterate on several things at once.
The convenience of list comprehensions encourages us to decompose a complicated task into a sequence of comprehensions. And to make that even more convenient (and faster/memory efficient) Python provides generators and all sort so itertools. In Python 3, range and dictionary keys/items/values all became generator like expressions.
This cannot be decided by your code that while loop worked faster than for loop. As of your question on when to use while and for loop, it is decided on the basis of whether we know the number of iterations. If number of iterations is known, for loop is preferred, whereas while loop is preferred when the iterations are indefinite.
for (int i : mynumber_list)
Iterator it = mynumber_list.iterator()
while (it.hasNext())
As can be seen from above, for loop is more readable, simple, and easy in traversing. Moreover, iterator.hasNext() has a very high probability of entering into an infinite loop
while can also be useful for user choice input questions, like keep executing the program until the user presses any other key except y. This is difficult to achieve with for loop.
This question already has answers here:
Removing Item From List - during iteration - what's wrong with this idiom? [duplicate]
(9 answers)
Closed 4 years ago.
I want to remove all even numbers in a list. But something confused me...
This is the code.
lst = [4,4,5,5]
for i in lst:
if i % 2 == 0:
print i
lst.remove(i)
print lst
It prints [4, 5, 5] Why not [5, 5]?
It should be like this
for i in lst[:]:
if i % 2 == 0:
print i
lst.remove(i)
print lst
Problem:
You are modifying the list while you iterate over it. Due to which the iteration is stopped before it could complete
Solution:
You could iterate over copy of the list
You could use list comprehension :
lst=[i for i in lst if i%2 != 0]
By using list.remove, you are modifying the list during the iteration. This breaks the iteration giving you unexpected results.
One solution is to create a new list using either filter or a list comprehension:
>>> filter(lambda i: i % 2 != 0, lst)
[5, 5]
>>> [i for i in lst if i % 2 != 0]
[5, 5]
You can assign either expression to lst if needed, but you can't avoid creating a new list object with these methods.
Other answers have already mentioned that you're modifying the list while iterating over it, and offered better ways to do it. Personally I prefer the list comprehension method:
odd_numbers = [item for item in numbers if item % 2 != 0]
For your specified case of a very small list, I would definitely go with that.
However, this does create a new list, which could be a problem if you have a very large list. In the case of integers, large probably means millions at least, but to be precise, it's however large it needs to be to start giving you issues with memory usage. In that case, here are a couple ways to do it.
One way is similar to the intent of the code in your question. You iterate over the list, removing the even numbers as you go. However, to avoid the problems that modifying a list you're iterating over can cause, you iterate over it backwards. There are ways to iterate forward, but this is simpler.
Here's one way using a while loop:
# A one hundred million item list that we don't want to copy
# even just the odd numbers from to put into a new list.
numbers = range(100000000) # list(range(100000000)) in Python 3
index = len(numbers) - 1 # Start on the index of the last item
while index >= 0:
if numbers[index] % 2 == 0:
numbers.pop(index)
index -= 1
Here's another way using a for loop:
# A one hundred million item list that we don't want to copy
# even just the odd numbers from to put into a new list.
numbers = range(100000000) # list(range(100000000)) in Python 3
for index in xrange(len(numbers) - 1, -1, -1): # range(...) in Python 3
if numbers[index] % 2 == 0:
numbers.pop(index)
Notice in both the while loop and for loop versions, I used numbers.pop(index), not numbers.remove(numbers[index]). First of all, .pop() is much more efficient because it provides the index, whereas .remove() would have to search the list for the first occurrence of the value. Second, notice that I said, "first occurrence of the value". That means that unless every item is unique, using .remove() would remove a different item than the one the loop is currently on, which would end up leaving the current item in the list.
There's one more solution I want to mention, for situations where you need to keep the original list, but don't want to use too much more memory to store a copy of the odd numbers. If you only want to iterate over the odd numbers once (or you're so averse to using memory that you'd rather recalculate things when you need to), you can use a generator. Doing so would let you iterate over the odd numbers in the list without needing any additional memory, apart from the inconsequential amount used by the generator mechanism.
A generator expression is defined exactly like a list comprehension, except that it's enclosed in parentheses instead of square brackets:
odd_numbers = (item for item in numbers if item % 2 != 0)
Remember that the generator expression is iterating over the original list, so changing the original list mid-iteration will give you the same problems as modifying a list while iterating over it in a for loop. In fact, the generator expression is itself using a for loop.
As an aside, generator expressions shouldn't be relegated only to very large lists; I use them whenever I don't need to calculate a whole list in one go.
Summary / TLDR:
The "best" way depends exactly what you're doing, but this should cover a lot of situations.
Assuming lists are either "small" or "large":
If your list is small, use the list comprehension (or even the generator expression if you can). If it's large, read on.
If you don't need the original list, use the while loop or for loop methods to remove the even numbers entirely (though using .pop(), not .remove()). If you do need the original list, read on.
If you're only iterating over the odd numbers once, use the generator expression. If you're iterating over them more than once, but you're willing to repeat computation to save memory, use the generator expression.
If you're iterating over the odd numbers too many times to recompute them each time, or you need random access, then use a list comprehension to make a new list with only the odd numbers in them. It's going to use a lot of memory, but them's the breaks.
As a general principle, you should not modify a collection while you are iterating over it. This leads to skipping of some elements, and index error in some cases.
Instead of removing elements from list, it would be easier if you just create another reference with same name. It has lesser time complexity too.
lst = filter(lambda i: i % 2 !=0, lst)
I have a string list
[str1, str2, str3.....]
and I also have a def to check the format of the strings, something like:
def CheckIP(strN):
if(formatCorrect(strN)):
return True
return False
Now I want to check every string in list, and of course I can use for to check one by one. But could I use map() to make code more readable...?
You can map your list to your function and then use all to check if it returns True for every item:
if all(map(CheckIP, list_of_strings)):
# All strings are good
Actually, it would be cleaner to just get rid of the CheckIP function and use formatCorrect directly:
if all(map(formatCorrect, list_of_strings)):
# All strings are good
Also, as an added bonus, all uses lazy-evaluation. Meaning, it only checks as many items as are necessary before returning a result.
Note however that a more common approach would be to use a generator expression instead of map:
if all(formatCorrect(x) for x in list_of_strings):
In my opinion, generator expressions are always better than map because:
They are slightly more readable.
They are just as fast if not faster than using map. Also, in Python 2.x, map creates a list object that is often unnecessary (wastes memory). Only in Python 3.x does map use lazy-computation like a generator expression.
They are more powerful. In addition to just mapping items to a function, generator expressions allow you to perform operations on each item as they are produced. For example:
sum(x * 2 for x in (1, 2, 3))
They are preferred by most Python programmers. Keeping with convention is important when programming because it eases maintenance and makes your code more understandable.
There is talk of removing functions like map, filter, etc. from a future version of the language. Though this is not set in stone, it has come up many times in the Python community.
Of course, if you are a fan of functional programming, there isn't much chance you'll agree to points one and four. :)
An example, how you could do:
in_str = ['str1', 'str2', 'str3', 'not']
in_str2 = ['str1', 'str2', 'str3']
def CheckIP(strN):
# different than yours, just to show example.
if 'str' in strN:
return True
else:
return False
print(all(map(CheckIP, in_str))) # gives false
print(all(map(CheckIP, in_str2))) # gives true
L = [str1, str2, str3.....]
answer = list(map(CheckIP, L))
answer is a list of booleans such that answer[i] is CheckIP(L[i]). If you want to further check if all of those values are True, you could use all:
all(answer)
This returns True if and only if all the values in answer are True. However, you may do this without listifying:
all(map(CheckIP, L)), as, in python3, `map` returns an iterator, not a list. This way, you don't waste space turning everything into a list. You also save on time, as the first `False` value makes `all` return `False`, stopping `map` from computing any remaining values
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.