This question already has answers here:
Accessing elements of Python dictionary by index
(11 answers)
Closed 4 years ago.
Iam trying to learn lambda expressions and came across the below :
funcs = {
idx: lambda: print(idx) for idx in range(4)
}
funcs[0]()
did not understand what is meant by funcs[0] ? Why the index and 0 ?
This is a cool example to show how binding and scope work with lambda functions. As Daniel said, funcs is a dictionary of functions with four elements in it (with keys 0,1,2 and 3). What is interesting is that you call the function at key 0, you expect the index (idx) that was used to create that key to then print out and you'll get 0. BUT I believe if you run this code, you actually print 3! WHAT!?!? So what is happening is that the lambda function is actually not binding the value of 0 when it stores the reference to the function. That means, when the loop finishes, the variable idx has 3 associated with it and therefore when you call the lambda function in position 0, it looks up the reference to the variable idx and finds the value 3 and then prints that out. Fun :)
That was referring more to your question title than to the test... The 0 is to access the function that has key 0 in the dictionary.
Related
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:
Getting one value from a tuple
(3 answers)
Closed 2 years ago.
variable1 = 1
variable2 = 2
a_tuple = (variable1, variable2)
def my_function(a):
pass
my_function(a_tuple(variable1))
Is there a way I can pass a specific value from a tuple into a function? This is a terrible example, but all I need to know is if I can pass variable1 from the tuple into the function, I understand in this instance I could just pass in variable 1, but its for more complicated functions that will get its data from a tuple, and I don't like the look of that many variables, too messy.
variable1 = 1
variable2 = 2
a_tuple = (variable1, variable2)
def my_function(a):
pass
my_function(*a_tuple)
This code would obviously provide an error as it unpacks the tuple and inserts 2 variables, to make this work in my program I would need a way to pass either variable1 or variable2 into the function. My question is can I define exactly which items from a tuple are passed into the function when calling the function? Latest version of Python if it matters.
P.S. I wrote print("hello world") for the first time 7 days ago, this is my first language and my first question I couldn't find an answer to. Go easy on me, and thank you for your time.
In the code you provided you don't have a tuple you have a list. But it is still pretty much the same.
In your example lets say that you wanted to pass the first variable you would do it like this:
my_function(a_tuple[0])
If you don't understand why there is a zero here and how does this work I highly suggest learning about lists before functions.
You just need to access individual elements of the tuple, using index notation:
my_function(a_tuple[0])
or
my_function(a_tuple[1])
You could, if you wanted, write a new function which takes a tuple and an index, and calls my_function with the appropriate element:
def my_other_function(tuple, index):
return my_function(tuple[index])
But I don't see how there would be much gain in doing that.
you can index a tuple or use the index method.
def my_function(a):
pass
my_function(a_tuple[0])
if you want to get the index of a value use the index() method
a_tuple.index(variable1) #this will return 0
This question already has answers here:
What do lambda function closures capture?
(7 answers)
Closed 3 years ago.
The following is a code snippet which I don't seem to get. The question is how to make the function output the desired result (not mentioning what the desired result is, I assume its printing 0 to 9).
Here is the question:
What does the below code snippet print out? How can we fix the anonymous functions to behave as we'd expect?
functions = []
for i in range(10):
functions.append(lambda : i)
for f in functions:
print(f())
In Python, no new scope will be produced in for loop
So after for i in range(10), the variable i is still exist, and its value == 9. And the lambda function lambda : i access the variable i
In order to output your desired result, you should pass the variable as a function argument in loop
functions = []
for i in range(10):
functions.append(lambda i=i: i)
for f in functions:
print(f())
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)
This question already has answers here:
How do I pass a variable by reference?
(39 answers)
Closed 7 years ago.
In detail, my question is this:
Given the following code,
x = 10
def func(x):
x = x+1
def main():
print(x)
func(x)
print(x)
if __name__ == '__main__':
main()
On running this I get:
10
10
Does this mean that Python does not pass values by reference?
And I did check through the other question of the sort, and most(if not all) included analogies of lists or other such examples.
Is it possible to explain this in simple terms, like just a simple integer?
Ps. I am a beginner to coding.
Thanks
If you are coming from a background such as C or C++, which I did, this can be maddening until you figure it out.
Python has names, not variables, and names are bound to objects. Effectively, you can think of all 'variables' or names, as being pointers to python objects.
In python, integers, floats, and strings are immutable. So when you do the following:
x = 10
x = x + 1
You are first binding the name x to the integer 10, then when you evaluate x + 1 you get a new object 11 and then you bind x to that object. Your x inside the function body is local to the function, and when you bind it to 11, the global x remains bound to 10.
If you were to pass a list to the function, and append something to the list, that list would be modified. A list in python is a mutable object. All names bound to the list would refer to the modified list.
As a result, when you pass mutable objects it may seem as if you are passing by reference, and when you pass immutable objects it may seem like you are passing by value.