I found this piece of code from a related question about reversing strings in python, but can someone please interpret it in plain English? Please note that I am still new to python and only learnt how to use while loops and functions yesterday :/ so I cant really put this into words myself cause my understanding isn't quite there yet.
anyways here is the code:
def reverse_string(string):
new_strings = []
index = len(string)
while index:
index -= 1
new_strings.append(string[index])
return ''.join(new_strings)
print(reverse_string('hello'))
Surely by knowing what it does, you can figure the code. In the while loop, the index value starts from the end of the string and counts down to 0. In each step, it adds that character (again, starting from the end) to the end of the list it is building. Finally, it combines the list into a string.
So, given 'abcd', the list gets built:
'abcd' #3 -> ['d']
'abcd' #2 -> ['d','c']
'abcd' #1 -> ['d','c','b']
'abcd' #0 -> ['d','c','b','a']
Well basically, the get the length of the string with the len method. Which will return you an integer value representing how long that string is.
They then use the length of this string and effectively iterate down to zero in a while loop. Using the -= operator.
Each iteration (meaning each time around the loop) it will take away from length until it reaches zero.
So lets use hello as an example input and go through this together.
reverse_string('hello') is how we would call the method, done in the print statement of your code.
We then enter the function and perform these steps:
We create a new empty array called new_strings.
We find the length of the initial string hello which returns us 5. Meaning that now index is equal to 5.
We create a while loop that keeps on going until index is no more using while(index): - a while loop like this treats a 0 value as falsy and will terminate upon reaching this. Therefore when index is 0 the loop will stop.
The first line of this loop performs index -= 1 which is the same as writing index = index - 1 so the very first loop through we get index = 5 - 1 and then now index is equal to 4.
Because Python then lets us access the character of a string using string[index] (and because this works from 0 -> n) performing hello[4] will in fact give us the character o.
We also append this character to the array new_strings meaning that as we go through the iterations to reach zero it will add each character backwards to this array giving us ['o', 'l', 'l', 'e', 'h']
Since index is now zero we leave the loop and perform a join operation on the array to yet again create a string. The command ''.join(new_strings) means that we wish to join the array we previously had without a separator. If we had done '#'.join(new_strings) we instead would have gotten o#l#l#e#h instead of olleh.
I hope this answer gives you some clarity.
Sure, It is very simple program. You should reffer string methods and string indexing in python to get clear idea. Let me explain this in deatial.
print(reverse_string('hello'))//The print function is calling another function
reverse_string and passing argument"hello".
def reverse_string(string):// The argument "hello" is stored in the variable
string in reverse_string function.
**new_strings = []** // created a new empty list
**index = len(string)** // len function returns the length of the argument
passed to the function. len("hello")=5 and assigned
to index. index=5.
while index: // while loop exicute until the condition get false .In this
example when index =0.in string the indexing start from 0 .For
example.
string[0]=h,string[1]=e,string[2]=l,string[3]=l,string[4]=o.
**index -= 1** //Note the last letter 'o' is in string[4] and the len
function returned 5 so we need to decrement index variable
to 4 so that it will pointing to string[4]=o
new_strings.append(string[index]) // append string[4] that is o and so on ...
return ''.join(new_strings)
Write a function that takes, as an argument, a list, identified by the variable aList. If the list only contains elements containing digits (either as strings as integers), return the string formed by concatenating all of the elements in the list (see the example that follows). Otherwise, return a string indicating the length of the list, as specified in the examples that follow.
I am just starting to learn how to code and this is my first CS class.
def amIDigits(aList):
for element in range(aList):
if element in aList.isdigit():
bList=[]
bList.append(aList)
return str(bList)
amIDigits([“hello”, 23]) should return the string “The length of the input is 2.”
amIDigits ([“10”, “111”]) should return the string “10111”
If I understand it right the output will be the joined digits even if they are not of the string format. So the best way is to use the all function (returns true if all elements of an iteration are true) and check if the string elements of the list are digits. If so, then return the join of all elements of the list converted to a string. Else, we return the length of the list using the new string formatting syntax (f represents string formatting and the {} return the result of an operation).
code:
def amIDigits(aList):
if all([str(i).isdigit() for i in aList]):
return ''.join(map(str,aList))
else:
return f'The length of the input is {len(aList)}.'
print(amIDigits(['hello', 23]))
print(amIDigits(['10', '111']))
print(amIDigits([55, 33]))
output:
The length of the input is 2.
10111
5533
First, I highly recommend having someone literally sit down and have you walk them through your thought process. It is more useful as a learner to debug your thought process than to have someone give you the answer.
One thing I noticed is that you created your empty list, bList, inside the for block. This will not work. You need to create an empty list to store things into before you begin for looping through the old list, otherwise you will be over-writing your new list every time it loops. So right now, your bList.append() statement is appending an element onto an empty list every time it runs. (You will get only the very last element in the aList stored into your bList.)
Another problem is that you use the range() function, but you don't need to. You want to look at each element inside the list. Range creates a sequence of numbers from 0 to whatever number is inside the parentheses: range() documentation. Your code tries to pass a list into range(), so it is invalid.
The "for blank in blank" statement breaks up whatever list is in the second blank and goes through each of its elements one at a time. For the duration of the for statement, the first blank is the name of the variable that refers to the element being looked at. so for example:
apples = ["Granny Smith","Red Delicious","Green"]
for apple in apples:
eat(apple) #yum!
The for in statement is more naturally spoken as "for each blank in blank:"
When I run the following code, it prints. However, I expected only one 1 rather than two.
for i in (1,1):
print(i)
Output
1
1
You are iterating over a tuple which contains two elements with value 1 so it prints 1 twice. Your code is equivalent to:
list = [1, 1]
for item in list:
print(item)
If you want to loop over a range of numbers:
for i in range(1, 2):
print(i)
Or if you want to print unique numbers or values in list or tuple convert it into the set it will automatically remove the duplicates
newList = set(list)
for value in newList:
print(value)
Sets and tuples are different. I suspect you are confusing them. On a set:
for i in {1, 1}:
print(i)
1
On a tuple:
for i in (1, 1):
print(i)
1
1
Think of sets as being like sets in math, and tuples as being more like sequences - you can have redundancies in a sequence, but not in a set.
After reading #KeshavGarg's answer, I suspect you thought that (a,b) in Python would mean stuff in a through b. As you're probably aware by now, this is not the case - you need range to get that. Interestingly (and I admit tangentially), the syntax we're discussing here varies by language. In MATLAB, the range syntax looks a lot more like what I assume you thought the Python range syntax was:
>> for i=1:4
disp(i)
end
There has been some discussion of implementing range literals (a la Matlab) in Python. This introduces a variety of interesting new problems, which you can read about in the documentation linked in the previous sentence.
For loops are always inclusive in Python: they always run over all elements of the iterator (other than exceptions such as break, etc.). What probably has you confused is the range syntax. range(1,1) will create a range object with one element. It is the range function, not the for-loop, that is exclusive, in the sense of not including the stop argument in the range object.
Hey I'm writing a program that receives a broadcast from Scratch and then determines based on the broadcast, where to proceed. The code turns the broadcast(list item) into a string and then breaks that string into a list using .split(). The only problem is the broadcast may only be 1 word instead of 2. Is there a way to check if one of the list items from .split() is empty and then change it to a place holder value?
Where I am having trouble
scratchbroadcast = str(msg[1])
BroadcastList = scratchbroadcast.split()
#starts the switch statement that interprets the message and proceeds
#to the appropriate action
v = BroadcastList[0]
w = BroadcastList[1]
if BroadcastList[1] == '':
w = "na"
If BroadcastList contains only one word then BroadcastList will be a single-element list, e.g.
>>> "foo".split()
['foo']
Obviously we can't check whether the second item in the list is an empty string ''; there isn't a second element. Instead, check the length of the list:
w = "na" if len(BroadcastList) == 1 else BroadcastList[1]
Alternatively, use try to catch the IndexError (it's easier to ask for forgiveness than permission):
try:
w = BroadcastList[1]
except IndexError:
w = "na"
Okay, first consider this: how about the third item? Or the fourth? Or the forty-second?
If the string doesn't contain a splitter character (e.g. a space), you wouldn't end up with a list of two items, one of which blank -- you would end up with a list of only one item.
In Python, the length of something is generally obtained through the built-in len() function:
len([]) # == 0
len(["foo"]) # == 1
len(["foo", "bar"]) # == 2
Therefore, you would do:
if len(broadcast_list) == 1:
broadcast_list += [""]
Other ways of doing the same thing include broadcast_list.append("") and broadcast_list.extend([""]). Which one to use is completely up to you; += and .extend are more or less equivalent while .append can only add a single element.
Looking at the rest of your code, your case calls won't work like you expect them to: in Python, strings are truthy, so 'string' or 'otherString' is basically the same as True or True. or is strictly a boolean operator and you can't use it for 'either this or that'.
Python is notorious for not having a switch statement. Your attempt at implementing one would actually be kind of cute had you gone through with it -- something like that can be a pretty good exercise in Python OOP and passing functions as first-class objects. (In my day-to-day use of Python I hardly ever need to do something like that, but it's great to have it in your conceptual toolkit.)
You might be happy to learn that Python strings have a lower method; with it, your code would end up looking something like this:
v = broadcast_list[0].lower()
if v == 'pilight':
# ...
else if v == 'motor':
# ...
else if v == 'camera':
# ....
On a side note, you might want to have a look a PEP8 which is the de facto standard for formatting Python code. If you want other people to be able to quickly figure out your code, you should conform at least to its most basic propositions - such as classes being CamelCased and variables in lowercase, rather than the other way around.
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.