This question already has answers here:
Why can't I iterate twice over the same iterator? How can I "reset" the iterator or reuse the data?
(5 answers)
Closed 4 years ago.
I am new to python and was learning its builtins but function all()
is not behaving as expected i don't know why.Here is my code
n=map(int,input().strip().split())
print(all([j>0 for j in n]))
print(list(n)) #this line returning empty list
Here are my inputs:
1 2 3 4 5 -9
And my output:
False
Does function all changes the original map object(values)? but something like this is not mentioned in given function definition on the docs link.
Thanks in advance
Map returns a generator object which you exhausted in your all function. Therefore when you call list on n, since n is empty/exhausted, it returns an empty list.
To fix just make n a list in the first place.
n=list(map(int,input().strip().split()))
print(all([j>0 for j in n]))
print(n)
Related
This question already has answers here:
How do I create a list with numbers between two values?
(12 answers)
Closed 17 days ago.
So I am defining a function that takes in one variable R. Then I need to create a list of all integers from 0 to R (in the context of the problem R will always be positive).
EX) When I do
R=5
print(list(0,R))
I just get a list with 2 elements: 0 and 5, but I want 0,1,2,3,4,5
The range() function returns a sequence of numbers, starting from 0 by default, and increments by 1 (by default), and stops before a specified number.
range(6) would return [0,1,2,3,4,5]
This question already has answers here:
How does a lambda function refer to its parameters in python?
(4 answers)
Closed 1 year ago.
I wrote a Python code like:
fun_list = []
for i in range(10):
fun_list.append(lambda : f(i))
for j in range(10):
fun_list[j]()
I want it to output numbers from 0 to 9, but actually it outputs 9 for ten times!
I think the question is that the variable is be transported into function f only it was be called. Once it was be called it will globally find variable named 'i'.
How to modify the code so that it can output numbers from 0 to 9?
Your Code doesn't make sensefun_list[j]() how are you calling a List Element?
If You Want to append the numbers mentioned above into your array then Correct Code is:-
fun_list = []
for i in range(10): ### the range has to be from 0 to 10 that is [0,10)
fun_list.append(i) ### You dont need lambda function to append i
What This does is also if of No need because you first of all havent initialized what is f?... Is it a function?. You here are not appending the value but instead appending the called function statement. You dont need that. Just do it simply like above....:-
fun_list.append(lambda : f(i))
This question already has answers here:
Understanding generators in Python
(13 answers)
Closed 1 year ago.
Here i want to generate 5 dicts, where 'a' is index 'i', 'b' is a random int between 0 and 5. But result stuck at 'a'=0, it kept generate new c from random.randint(0,5) and 'i' remained the same. how to fix this? thx a lot
def wdg():
for i in range(5):
c = random.randint(0,5)
yield {'a':i,'b':c}
next(wdg())
Assign generator to a variable for example 'generator'. Then each time you call next(generator) one execution will occur, leading to increment i by one and generate random c:
generator = wdg()
print(next(generator))
This question already has answers here:
calculate length of list recursively [duplicate]
(2 answers)
Closed 7 years ago.
So, my goal is to count the number of arguments in a list using recursion. However, as I'm only to use one argument in the function call, I don't know how to solve it without a second "count" argument.
So far I have this, where I accumulate the 1's together.
def countElements(a):
if a==[]:
return []
else:
return [1] + countElements(a[1:])
def main():
a=[3,2,5,3]
print(countElements(a))
main()
Instead of returning an empty list in the base case, return 0, and instead of [1] + countElements(a[1:]), return 1 + countElements(a[1:]). That way, you're just keeping the running count instead of getting a list back.
This question already has answers here:
Why can't I iterate twice over the same iterator? How can I "reset" the iterator or reuse the data?
(5 answers)
Closed 9 years ago.
I wish to iterate over a list_iterator twice. When I currently try to do this, the iterator has nothing to iterate over the second time. Can I reset it?
l = iter(["1","2","3","4"])
for i in l:
print(i)
for i in l:
print(i)
A list_iter object is being passed to the function in which I wish to iterate twice in. Is it bad practive to pass around list_iterator objects?
You could always use itertools.tee to get copies of the iterable.
Ala
l = iter(["1","2","3","4"])
# Don't use original....
original, new = tee(l)