List Comprehension Python Prime numbers - python

I came across a solution on Stack Overflow to generate prime numbers using list comprehension. But was unable to understand what does the inner for loop do.
I have tried something like
[x for x in range(5,20) for y in range(2,int(x/2)+1) if any(x%y == 0)]
It throws an error: 'bool' object is not iterable
I know that my syntax is wrong but logically for prime numbers we have a for loop followed by a for loop and then a if condition to calculate remainder(x%y).
But the answer on Stack Overflow is
[x for x in range(2, 20) if all(x % y != 0 for y in range(2, x))]
I understood the reason why all is used, but I am unable to get how the condition inside all() is working as ideally for should be following if so that range(2,x) is iterated and y gets values which are in turn used for computing(x%y). How can y be used even before it is has been assigned a value.

That is just the wonderful thing about list comprehension if it can work normally like the for loop, people wont create it because the for loop is more readable and understandable.
You may find out that the result of list comprehension is always a list, meanwhile the result of for loop would always many single values and these single values is a part of iterable
[x +1 for x in range(1,5)]
[2, 3, 4, 5]
for x in range (1,10): print(x+1)
2
3
4
5
You can simply understand that the loop comprehension already have the list of values, then they just simply feed orderly to the condition value by value. Like this:
[1+1 , 2+1 , 3+1 , 4+1]
Your code is wrong because you inherit too much from the ordinary for loop. Your code written in for loop would be like this:
for x in range(5,20):
for y in range(2,int(x/2)+1):
if any(x%y == 0):
print(x)
And the result would obviously:
TypeError: 'bool' object is not iterable
because any requires an iterable such as a generator expression or a **list** as mentioned above by #meowgoesthedog . Coincidentally, list is just all about list comprehension. However, you need comprehend it in order to utilize the list comprehension well. It sometimes happens to me too, in your case, the for y in range(2,int(x/2)+1) works as a normal for loop.
This is the syntax of list comprehension.
In side the condition if which is optional predicate. We can create another list comprehension by following the rules with x%y==0 is output expression and a variable y representing members of the input sequence range(2,int(x/2)+1)

all() and any() works on itterable objects. For example all([True, True, False, True]) returns False. You cant use any(True) (like in your example: any(x%y == 0))
This statement [x for x in range(2, 20) if all(x % y != 0 for y in range(2, x))] can be translated to this code:
res = []
for x in range(2, 20):
temporary_list = (x%y != 0 for y in range(2,x))
if all(temporary_list):
res.append(x)
Ps. I saw in comments that you are not sure how y is declared. In python, there are more great structures than list of comprehension. One of them is generator of comprehension - I believe it is used in this case.

The syntax all and any work on iterable objects (list, sets, etc). Therefore you get an error when you apply it on boolean - x%y==0.
You can use any in the following manner -
[x for x in range(5,20) if not any([x % y == 0 for y in range(2, int(x/2)+1)])]
or -
[x for x in range(2, 20) if not any(x % y == 0 for y in range(2, int(x/2)+1))]
As any and all complement each other.

Related

what's the difference between filter and comprehention with if?

def anagramwordchecker(z,w):
if sorted([x for x in w])==sorted([x for x in z]):return True
return False
def anagramlistchecker(l,w):
d={}
for x in w:
d.update({x:w.count(x)})
l=list(filter(lambda x:anagramwordchecker(x,w),l))
return l
print(anagramlistchecker(['bcda', 'abce', 'cbda', 'cbea', 'adcb'],'abcd'))
trying to check which words are anagram.
using both of this it will print the same:
l=[x for x in l if anagramwordchecker(x,w)]
l=list(filter(lambda x:anagramwordchecker(x,w),l))
and it will be:
['bcda', 'cbda', 'adcb']
then what's the difference? any advantage using filter? cause comprehension is easier.
If you print the results of the following example, you will know which one is faster (Comments are results I got).
timeit.Timer('''[x for x in range(100) if x % 2 == 0]''' ).timeit(number=100000)
timeit.Timer('''list(filter(lambda x: x % 2 == 0, range(100)))''').timeit(number=100000)
# 0.3664856200000486
# 0.6642515319999802
So in your case, list comprehension would be faster. But let's see the following example.
timeit.Timer('''[x for x in range(100) if x % 2 == 0]''' ).timeit(number=100000)
timeit.Timer('''(x for x in range(100) if x % 2 == 0)''' ).timeit(number=100000)
timeit.Timer('''filter(lambda x: x % 2 == 0, range(100))''').timeit(number=100000)
# 0.5541256509999357
# 0.024836917000016
# 0.017953075000036733
The results show that casting an iterable to list takes much time and filter is faster than generator expression. So if your result does not really have to be a list, returning an iterable in a timely manner would be better.
As stated in here,
Note that filter(function, iterable) is equivalent to the generator expression (item for item in iterable if function(item)) if function is not None and (item for item in iterable if item) if function is None.
But list comprehension can do much more than simply filtering. If filter is given to the interpreter, it will knows it is a filter function. However, if a list comprehension is given to the interpreter, the interpreter does not know what it really is. After taking some time interpreting the list comprehension to something like a function, it would be a filter or filterfalse function in the end. Or, something else completely different.
filter with not condition can do what filterfalse does. But filterfalse is still there. Why? not operator does not need to be applied.
There is no magic. Human-friendly 1-for-many grammars are based on encapsulation. For them to be machine-executable binaries, they need to be decapsulated back and it takes time.
Go with a specific solution if it is enough than taking a more general solutions. Not only in coding, general solutions are usually for convenience, not for best results.

Is it possible to limit the length of a conditional list comprehension in Python?

I am doing a conditional list comprehension e.g. newlist = [x for x in list if x % 2 == 0]. I want to limit the length of the resulting list to a specific number.
Is this possible without first comprehending the entire list and then slicing it?
I imagine something that has the functionality of:
limit = 10
newlist = []
for x in list:
if len(newlist) > limit:
break
if x % 2 == 0:
newlist.append(x)
Slicing the original list (e.g. [x for x in list[:25] if x % 2 == 0] is not possible, as the if condition does not return True in any predictable intervals in my specific use case.
Many thanks in advance.
Please don't name any variables list as it shadows the built-in list constructor. I used li as a replacement for the input list here.
import itertools as it
gen = (x for x in li if x % 2 == 0) # Lazy generator.
result = list(it.islice(gen, 25))
Since you are creating a list with the list comprehension you can slice your list directly after it is created.
[x for x in list[:25] if x % 2 == 0][:limit]

Stacking inline For plus If else

I wrote a line that should sum up all numbers of a string except 0. On a zero it should add 5.
Whats wrong with
s="123450"
o=sum([int(x) for x in s if int(x) != 0 else 5])
it gives syntax error, but
s="123450"
o=sum([int(x) for x in s if int(x) != 0])
works fine.
if at the end of a list comprehension is used as a filter. It can only cause values to be dropped, not replace them with other things. To map values, you have to move the if/else earlier in the comprehension:
o=sum([int(x) if int(x) != 0 else 5 for x in s])
The correct syntax for the first case if -
s="123450"
o=sum([int(x) if int(x) != 0 else 5 for x in s ])
print(o)
OUTPUT :
20
There are many good answers already available on StackOverflow over if-else in list comprehension. Check these -
if else in a list comprehension
List comprehension with if statement
if/else in a list comprehension
More of an aside really (as the general issue of how you are using if inside list comprehensions is well covered in other answers), in this particular case you could do the following:
o = sum(int(x) or 5 for x in s)
This works because or will use the first value if it is "true" (which in the case of integers means non-zero) or the second value if the first is "false" (here, 0).
The other difference here is that I've used a generator expression instead of a list comprehension -- there is no need to build a list just in order to sum the values.

Difference In List - Python - Syntax Explanation [duplicate]

This question already has answers here:
Explanation of how nested list comprehension works?
(11 answers)
Closed 6 years ago.
Can someone please explain the meaning the syntax behind the following line of code:
temp3 = [x for x in temp1 if x not in s]
I understand it's for finding the differences between 2 lists, but what does the 'x' represent here? Each individual element in the list that is being compared? I understand that temp1 and s are lists. Also, does x for x have to have the same variable or could it be x for y?
[x for x in temp1 if x not in s]
It may help to re-order it slightly, so you can read the whole thing left to right. Let's move the first x to the end.
[for x in temp1 if x not in s yield x]
I've added a fake yield keyword so it reads naturally as English. If we then add some colons it becomes even clearer.
[for x in temp1: if x not in s: yield x]
Really, this is the order that things get evaluated in. The x variable comes from the for loop, that's why you can refer to it in the if and yield clauses. But the way list comprehensions are written is to put the value being yielding at the front. So you end up using a variable name that's not yet defined.
In fact, this final rewrite is exactly how you'd write an explicit generator function.
def func(temp1, s):
for x in temp1:
if x not in s:
yield x
If you call func(temp1, s) you get a generator equivalent to the list. You could turn it into that list with list(func(temp1, s)).
It iterates through each element in temp1 and checks to see if it is not in s before including it in temp3.
It is a shorter and more pythonic way of writing
temp3 = []
for item in temp1:
if item not in s:
temp3.append(item)
Where temp1 and s are the two lists you are comparing.
As for your second question, x for y will work, but probably not in the way you intend to, and certainly not in a very useful way. It will assign each item in temp1 to the variable name y, and then search for x in the scope outside of the list comprehension. Assuming x is defined previously (otherwise you will get NameError or something similar), the condition if x not in s will evaluate to the same thing for every item in temp1, which is why it’s not terribly useful. And if that condition is true, your resulting temp3 will be populated with xs; the y values are unused.
Do not take this as saying that using different variables in a list comprehension is never useful. In fact list comprehensions like [a if condition(x) else b for x in original_sequence] are often very useful. A list comprehension like [a for x in original_sequence if condition(x)] can also be useful for constructing a list containing exactly as many instances of a as the number of items in original_sequence that satisfy condition().
Try yourself:
arr = [1,2,3]
[x+5 for x in arr]
This should give you [6, 7, 8] that are the values on the [1,2,3] list plus 5. This syntax is know as list comprehension (or mapping). It applies the same instructions to all elements on a list. Would be the same as doing this:
for x in arr:
arr += 5
X is same variable and it is not y. It works same as below code
newList = []
for x in temp1:
if x not in s:
newList.append(x)
So x for x, here first is x which is inside append in code and x after for is same as for x in temp1.

Removing Redundancies/Condensing Code

Both code examples below are old examples of a problem I have where I am iterating through a list of numbers to find numbers that fit a list of conditions but couldn't find a neat way to express it except for:
condition1 and condition2 and condition3 and condition4 and condition5
Two examples of the above:
if str(x).count("2")==0 and str(x).count("4")==0 and str(x).count("6")==0 and str(x).count("8")==0 and str(x).count("0")==0:
if x % 11==0 and x % 12 ==0 and x % 13 ==0 and x%14==0 and x %15 == 0 and x%16==0 and x%17==0 and x%18==0 and x%19==0 and x%20==0:
Is there a simple, more neat way of doing this?
My first retry was to store the conditions in a list like below:
list=["2","4","6","8","0"]
for element in list:
#call the function on all elements of the list
list=[11,12,13,14,15,16,17,18,19,20]
for element in list:
#call the function on all elements of the list
but I was hoping for an even neater/simpler way.
You can use a generator expression like this
def f(n):
return x%n
if all(f(element) for element in lst):
...
If the function/calculation is not too complex, you can just put it inline
if all(x % element for element in lst):
...
The built in function all can simplify this, if you can express your conditions in a generator expression:
result = all(x % n == 0 for n in xrange(11, 21))
It returns a boolean result indicating whether or not all the elements of its iterable argument are True, short-circuiting to end evaluation as soon as one is False.
This is the second question I've seen in the last hour or so for which all is the answer - must be something in the air.

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