List Comprehensions to replace element - list-comprehension

for i in range(1, len(A)):
A[i] = A[i-1] + A[i]

You can't do that with a list comprehension as they don't allow assignments.
You can use a simple generator function:
def func(lis):
yield lis[0]
for i,x in enumerate(lis[1:],1):
lis[i] = lis[i-1] + x
yield lis[i]
>>> A = [1, 2, 3, 4, 5, 6, 7]
>>> list(func(A))
[1, 3, 6, 10, 15, 21, 28]

though less efficient, this does give the desired output. But I think I'm getting closer to O(n**2) on this one.
A = [sum(A[:i+1]) for i, _ in enumerate(A)]
afaik this can't be done with a list comprehension the way you want to. I would suggest using the for loop version you've provided. Even if it was possible with a list comprehension, there's no point when you can just modify the list in place.

Use B (another temporary variable).

This should do the trick.
def func(L):
it = iter(L)
v = it.next()
yield v
for x in it:
v += x
yield v
A = [1, 2, 3, 4, 5, 6, 7]
print list(func(A))
This creates an iterator that returns one value at the time. To get the full new list at once you need to use a list() call around the function call, like:
list(func(A))
This generator function should work on any iterable (also those that don't support getting value based on index, like L[0])
I don't think there's an efficient way to do this with a comprehension list.

A one-line solution use reduce:
>>> the_list = [1,2,3,4,5]
>>> reduce(lambda result, x: result+[result[-1] + x] ,the_list, [0])[1:]
[1, 3, 6, 10, 15]

Related

Python} Reversing List Without Using String Method

How should I write the code with the following problem?
Implement the function reverse_print(lst) that prints out the contents of the given list ‘lst’
in reverse order. For example, given a list [3, 6, 2, 1], the output should be 1, 2, 6, 3 (vertical
printout allowed). For this code, you are only allowed to use a single for-loop. Without String Method
Also Not using Print[::-1]
Assuming you are not allowed to just call list.reverse() you can use range with a negative step to iterate the list in reverse order:
def reverse_print(lst):
out = []
for i in range(len(lst) - 1, -1, -1):
out.append(lst[i])
print(*out, sep=", ")
inp = [1, 2, 3, 4]
reverse_print(inp)
Output: 4, 3, 2, 1
You may try something like this
def reverse_print(lst):
rev = [lst[abs(i-l)-1] for i in range(l)]
return rev
lst = [3,6,2,1]
l = len(lst)
print(reverse_print(lst))

How do I multiply elements in a list and find the sum in a simple way without numpy, zip etc

I have to multiply the same index of two lists and then find the sum of that.
Please help me out! Thank you.
Try this,
>>> A=[2, 3, -6, 7, 10, 11]
>>> B=[1, 2, 3, 4, 5, 6]
>>> sum([x * y for x, y in zip(A, B)])
134
Let me explain what I did in my answer, I used zip() python Built-in Function and this is what documentation mention about it.
Make an iterator that aggregates elements from each of the iterables.
Returns an iterator of tuples, where the i-th tuple contains the i-th element from each of the argument sequences or iterables. The iterator stops when the shortest input iterable is exhausted. With a single iterable argument, it returns an iterator of 1-tuples. With no arguments, it returns an empty iterator.
Little bit confusing, right? Check the below example:-
>>> A=[2, 3, -6, 7, 10, 11]
>>> B=[1, 2, 3, 4, 5, 6]
>>> zip(A,B)
<zip at 0x1fde73e6f88> # it display something like this (zip object)
>>> list(zip(A,B)) # for visualization purpose, convert zip object to list
[(2, 1), (3, 2), (-6, 3), (7, 4), (10, 5), (11, 6)]
I think you can get clear idea what happen inside zip() function. Then multiply each and every value in zip object using python List Comprehensions to answer your question more pythonic. So zip object now created for us a new value series using A and B list. We assigned those values for x and y, multiply them and save in list.
>>> [x * y for x, y in zip(A, B)]
[2, 6, -18, 28, 50, 66]
After all the steps we used sum() to calculate the Sum a list of numbers in Python.
>>> sum([2, 6, -18, 28, 50, 66])
134
That's all, if you didn't get anything please add a comment to this answer.
AB = [A[i] * B[i] for i in range(len(A))]
sum(AB)
Alternatively, try
AB = [value_ * B[i] for i, value_ in enumerate(A)]
sum(AB)
I believe you want this:
def lists(A,B):
C = 0
for i in range(len(A)):
C += (A[i] * B[i])
return C
Now you can call your method lists with lists A and B like this:
A=[2, 3, -6, 7, 10, 11]
B=[1, 2, 3, 4, 5, 6]
lists(A,B)
Which will return 134. Your code was wrong because of your indentation. You had put your return statement inside the for loop, so your code would return C value in the first iteration, which was 0 + 2*1.
with list comprehension:
A = [2, 3, -6, 7, 10, 11]
B = [1, 2, 3, 4, 5, 6]
print (sum([A[i]*B[i] for i in range(len(A))]))
output:
134
your code:
def lists(A, B):
C = 0
for i in range(len(A)):
C += (A[i] * B[i])
return C # <-----
A = [2, 3, -6, 7, 10, 11]
B = [1, 2, 3, 4, 5, 6]
print (lists(A,B))
NOTE: you need to put your return statement out of the for loop.
If return statement is reached during the execution of a function, it will exit from function, what does it mean in your case if you have return in your for loop, in first iteration return function will be reached and exit (you got result 2 because in first iteration yuo have 2*1)

How to replace every element in a given list to its square and not create a new list with the squares

I want to make a function that takes a list of integers and replaces the elements in the list with their respective squares.
I tried reassigning every element by virtue of its position (index) in the list, but for some reason the second element in the list gets squared twice.
def square_list(list1):
for i in list1:
list1[list1.index(i)] = i**2
print(list1)
square_list([1, 2, 3, 4, 5])
I expect the printed list to be [1, 4, 9, 16, 25] since the list I'm testing the function with is [1, 2, 3, 4, 5].
If a function is required to square list elements in-place, use it:
def square_list(list_1):
for i in range(len(list_1)):
list_1[i] = list_1[i]**2
my_list = [1, 2, 3, 4, 5]
square_list(my_list)
print(my_list)
Since the function doesn't return anything, square_list([1, 2, 3, 4, 5]) is useless.
Python's map built-in function is good for this, so you don't really have to write your own function.
l = [1, 2, 3, 4, 5]
l = list(map(lambda x: x**2, l))

To make odd position in list go forward for one step

I found this code from stackoverflow ... and wondering how can I move index position that I want.
I tried to use for loop and [::1]. And by making, len(a)*[0]...I couldn't make it.
Is there any way to fix items on its position in list?
Second, without using method below, is there another way to reorder items in list?
'''
mylist=['a','b','c','d','e']
myorder=[3,2,0,1,4]
'''
a = [1,2,3,4,5,6,7]
b = ((a+a[:0:-1])*len(a))[::len(a)][:len(a)]
[1, 7, 2, 6, 3, 5, 4] <=b
[7, 1, 6, 2, 5, 3, 4] <= the result i want
Thanks in advance.
I'm not sure if this is what you want:
someList = [1,2,3,4,5,6,7]
orderedList = sorted(someList)
reversedOrderedList = orderedList[::-1]
finalList = []
for i in range(len(someList)):
if i % 2 == 0:
finalList.append(reversedOrderedList[i//2])
else:
finalList.append(orderedList[i//2])
print(finalList)
output:
[7, 1, 6, 2, 5, 3, 4]
if so, you can write it in shorter way (without reversedOrderedList):
someList = [1,2,3,4,5,6,7]
orderedList = sorted(someList)
finalList = []
for i in range(len(someList)):
if i % 2 == 0:
finalList.append(orderedList[-1-i//2])
else:
finalList.append(orderedList[i//2])
print(finalList)
and from here you can write it without if statement:
someList = [1,2,3,4,5,6,7]
orderedList = sorted(someList)
for i in range(len(someList)):
finalList.append(orderedList[((-1)**(i%2+1)-1)//2 + ((- 1)**(i%2+1))*(i//2)])
print(finalList)
It is not pretty but after that you can easily write a generator.
Zip the list with its reversed version, flatten it and take the first half:
from itertools import chain
a = [1,2,3,4,5,6,7]
b = list(chain.from_iterable(zip(a[::-1], a)))
print(b[:len(b) // 2])
Output
[7, 1, 6, 2, 5, 3, 4]

Difference between normal for loop statement and for loop statement inside a list initialization in python [duplicate]

I have the following code:
[x ** 2 for x in range(10)]
When I run it in the Python shell, it returns:
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
I've searched and it seems this is called a list comprehension and similarly there seem to be set/dict comprehensions and generator expressions. But how does it work?
From the documentation:
List comprehensions provide a concise way to create lists. Common applications are to make new lists where each element is the result of some operations applied to each member of another sequence or iterable, or to create a subsequence of those elements that satisfy a certain condition.
About your question, the list comprehension does the same thing as the following "plain" Python code:
>>> l = []
>>> for x in range(10):
... l.append(x**2)
>>> l
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
How do you write it in one line? Hmm...we can...probably...use map() with lambda:
>>> list(map(lambda x: x**2, range(10)))
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
But isn't it clearer and simpler to just use a list comprehension?
>>> [x**2 for x in range(10)]
[0, 1, 4, 9, 16, 25, 36, 49, 64, 81]
Basically, we can do anything with x. Not only x**2. For example, run a method of x:
>>> [x.strip() for x in ('foo\n', 'bar\n', 'baz\n')]
['foo', 'bar', 'baz']
Or use x as another function's argument:
>>> [int(x) for x in ('1', '2', '3')]
[1, 2, 3]
We can also, for example, use x as the key of a dict object. Let's see:
>>> d = {'foo': '10', 'bar': '20', 'baz': '30'}
>>> [d[x] for x in ['foo', 'baz']]
['10', '30']
How about a combination?
>>> d = {'foo': '10', 'bar': '20', 'baz': '30'}
>>> [int(d[x].rstrip('0')) for x in ['foo', 'baz']]
[1, 3]
And so on.
You can also use if or if...else in a list comprehension. For example, you only want odd numbers in range(10). You can do:
>>> l = []
>>> for x in range(10):
... if x%2:
... l.append(x)
>>> l
[1, 3, 5, 7, 9]
Ah that's too complex. What about the following version?
>>> [x for x in range(10) if x%2]
[1, 3, 5, 7, 9]
To use an if...else ternary expression, you need put the if ... else ... after x, not after range(10):
>>> [i if i%2 != 0 else None for i in range(10)]
[None, 1, None, 3, None, 5, None, 7, None, 9]
Have you heard about nested list comprehension? You can put two or more fors in one list comprehension. For example:
>>> [i for x in [[1, 2, 3], [4, 5, 6]] for i in x]
[1, 2, 3, 4, 5, 6]
>>> [j for x in [[[1, 2], [3]], [[4, 5], [6]]] for i in x for j in i]
[1, 2, 3, 4, 5, 6]
Let's talk about the first part, for x in [[1, 2, 3], [4, 5, 6]] which gives [1, 2, 3] and [4, 5, 6]. Then, for i in x gives 1, 2, 3 and 4, 5, 6.
Warning: You always need put for x in [[1, 2, 3], [4, 5, 6]] before for i in x:
>>> [j for j in x for x in [[1, 2, 3], [4, 5, 6]]]
Traceback (most recent call last):
File "<input>", line 1, in <module>
NameError: name 'x' is not defined
We also have set comprehensions, dict comprehensions, and generator expressions.
set comprehensions and list comprehensions are basically the same, but the former returns a set instead of a list:
>>> {x for x in [1, 1, 2, 3, 3, 1]}
{1, 2, 3}
It's the same as:
>>> set([i for i in [1, 1, 2, 3, 3, 1]])
{1, 2, 3}
A dict comprehension looks like a set comprehension, but it uses {key: value for key, value in ...} or {i: i for i in ...} instead of {i for i in ...}.
For example:
>>> {i: i**2 for i in range(5)}
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}
And it equals:
>>> d = {}
>>> for i in range(5):
... d[i] = i**2
>>> d
{0: 0, 1: 1, 2: 4, 3: 9, 4: 16}
Does (i for i in range(5)) give a tuple? No!, it's a generator expression. Which returns a generator:
>>> (i for i in range(5))
<generator object <genexpr> at 0x7f52703fbca8>
It's the same as:
>>> def gen():
... for i in range(5):
... yield i
>>> gen()
<generator object gen at 0x7f5270380db0>
And you can use it as a generator:
>>> gen = (i for i in range(5))
>>> next(gen)
0
>>> next(gen)
1
>>> list(gen)
[2, 3, 4]
>>> next(gen)
Traceback (most recent call last):
File "<input>", line 1, in <module>
StopIteration
Note: If you use a list comprehension inside a function, you don't need the [] if that function could loop over a generator. For example, sum():
>>> sum(i**2 for i in range(5))
30
Related (about generators): Understanding Generators in Python.
There are list, dictionary, and set comprehensions, but no tuple comprehensions (though do explore "generator expressions").
They address the problem that traditional loops in Python are statements (don't return anything) not expressions which return a value.
They are not the solution to every problem and can be rewritten as traditional loops. They become awkward when state needs to be maintained & updated between iterations.
They typically consist of:
[<output expr> <loop expr <input expr>> <optional predicate expr>]
but can be twisted in lots of interesting and bizarre ways.
They can be analogous to the traditional map() and filter() operations which still exist in Python and continue to be used.
When done well, they have a high satisfaction quotient.
If you prefer a more visual way of figuring out what's going on then maybe this will help:
# for the example in the question...
y = []
for x in range(10):
y += [x**2]
# is equivalent to...
y = [x**2 for x in range(10)]
# for a slightly more complex example, it is useful
# to visualize where the various x's end up...
a = [1,2,3,4]
b = [3,4,5,6]
c = []
for x in a:
if x in b:
c += [x]
# \ \ /
# \ _____\______/
# \ / \
# \/ \
# /\ \
# / \ \
# / \ \
c = [x for x in a if x in b]
print(c)
...produces the output [3, 4]
I've seen a lot of confusion lately (on other SO questions and from coworkers) about how list comprehensions work. A wee bit of math education can help with why the syntax is like this, and what list comprehensions really mean.
The syntax
It's best to think of list comprehensions as predicates over a set/collection, like we would in mathematics by using set builder notation. The notation actually feels pretty natural to me, because I hold an undergrad degree in Mathematics. But forget about me, Guido van Rossum (inventor of Python) holds a masters in Mathematics and has a math background.
Set builder notation crash course
Here's the (very basics) of how set builder notation works:
So, this set builder notation represents the set of numbers that are strictly positive (i.e. [1,2,3,4,...]).
Points of confusion
1) The predicate filter in set builder notation only specifies which items we want to keep, and list comprehension predicates do the same thing. You don't have to include special logic for omitting items, they are omitted unless included by the predicate. The empty predicate (i.e. no conditional at the end) includes all items in the given collection.
2) The predicate filter in set builder notation goes at the end, and similarly in list comprehensions. (some) Beginners think something like [x < 5 for x in range(10)] will give them the list [0,1,2,3,4], when in fact it outputs [True, True, True, True, True, False, False, False, False, False]. We get the output [True, True, True, True, True, False, False, False, False, False] because we asked Python to evaluate x < 5 for all items in range(10). No predicate implies that we get everything from the set (just like in set builder notation).
If you keep set builder notation in the back of your mind while using list comprehensions, they're a bit easier to swallow.
HTH!
Introduction
A list comprehension is a high level, declarative way to create a list in Python. The main benefits of comprehensions are readability and maintainability. A lot of people find them very readable, and even developers who have never seen them before can usually guess correctly what it means.
# Snippet 1
squares = [n ** 2 for n in range(5)]
# Snippet 2
squares = []
for n in range(5):
squares.append(n ** 2)
Both snippets of code will produce squares to be equal to [0, 1, 4, 9, 16].
Notice that in the first snippet, what you type is declaring what kind of list you want, while the second is specifying how to create it. This is why a comprehension is a high-level and declarative.
Syntax
[EXPRESSION for VARIABLE in SEQUENCE]
EXPRESSION is any Python expression, but it is typical to have some variable in it. This variable is stated in VARIABLE field. SEQUENCE defines the source of values the variable enumerates through.
Considering Snippet 1, [n ** 2 for n in range(5)]:
EXPRESSION is n ** 2
VARIABLE is n
SEQUENCE is range(5)
Notice that if you check the type of squares you will get that the list comprehension is just a regular list:
>>> type(squares)
<class 'list'>
More about EXPRESSION
The expression can be anything that reduces to a value:
Arithmetic expressions such as n ** 2 + 3 * n + 1
A function call like f(n) using n as variable
A slice operation like s[::-1]
Method calls bar.foo()
...
Some examples:
>>> [2 * x + 3 for x in range(5)]
[3, 5, 7, 9, 11]
>>> [abs(num) for num in range(-5, 5)]
[5, 4, 3, 2, 1, 0, 1, 2, 3, 4]
>>> animals = ['dog', 'cat', 'lion', 'tiger']
>>> [animal.upper() for animal in animals]
['DOG', 'CAT', 'LION', 'TIGER']
Filtering:
The order of elements in the final list is determined by the order of SEQUENCE. However, you can filter out elements adding an if clause:
[EXPRESSION for VARIABLE in SEQUENCE if CONDITION]
CONDITION is an expression that evaluates to True or False. Technically, the condition doesn't have to depend upon VARIABLE, but it typically uses it.
Examples:
>>> [n ** 2 for n in range(5) if n % 2 == 0]
[0, 4, 16]
>>> animals = ['dog', 'cat', 'lion', 'tiger']
>>> [animal for animal in animals if len(animal) == 3]
['dog', 'cat']
Also, remember that Python allows you to write other kinds of comprehensions other than lists:
dictionary comprehensions
set comprehensions

Resources