Is there any way to import an argument from a function as a variable to another function?
For example:
untitled1.py
def fun1(a,b):
x = a+b
return x
and I want to pass the a as a variable to the following function which is another py file..
untitled2.py
def fun2(c):
d=c/a
return d
You should define the variable 'a' as global.
variable.py
class A:
value = None
untitled1.py
def fun1(a,b):
x = a.value+b
return x
untitled2.py
def fun2(c):
d=c/a.value
return d
where to call func1, func2
A.value = 2
func1(A, 4)
func2(10)
Try this:
`untitled1.py
def fun1(a,b):
x = a+b
return x,a
untitled2.py
import untitled1
def fun2(c):
var1,var2 = fun1(a,b)
d=c/var2
return d`
Related
In the code below, I was getting this.
NameError: name 'digitsum' is not defined
Please note that we can return values but that's not I want exactly here. I just want to change the outside values inside function f.
class Solution:
def compute(n) -> int:
digitsum = 0
def f(k):
global digitsum
if k==0:
return
digitsum+=k%10
f(k//10)
f(3648)
return digitsum
In this case digitsum is not a global variable, which is why you're getting a NameError. In this particular case what you want instead is nonlocal.
A simplified example:
>>> def foo():
... x = 0
... def f():
... nonlocal x
... x += 1
... f()
... print(x)
...
>>> foo()
1
class Solution:
digitsum = 0
def compute(self,n) -> int:
self.digitsum = 0
def f(k):
if k==0:
return
self.digitsum+=k%10
f(k//10)
f(n)
return self.digitsum
Solution().compute(45)
def MainCount(f):
def progFirst(*args,**kwargs):
progFirst.calls+=1
return f(*args,**kwargs)
progFirst.calls=0
return progFirst
#MainCount
def progSecond(i):
return i+1
#MainCount
def Count(i=0,j=1):
return i*j+1
print(progSecond.calls)
for n in range(5):
progSecond(n)
Count(j=0,i=1)
print(Count.calls)
Output :0
1
As per my understanding MainCount(probSecond) but I am not understant then how probSecond.calls equal to zero same in Count.calls also
As You Can See in MainCount function probFirst.Calls is attribute of function .When MainCount(probSecond) Now probSecond.calls is also attribute of MainCount function.
# A Python example to demonstrate that
# decorators can be useful attach data
# A decorator function to attach
# data to func
def attach_data(func):
func.data = 3
return func
#attach_data
def add (x, y):
return x + y
# Driver code
# This call is equivalent to attach_data()
# with add() as parameter
print(add(2, 3))
print(add.data)
How can i get the output from another script?
My first script to run:
from test2 import *
class Test():
def todo (self):
mult()
addx()
if __name__ == '__main__':
Test().todo()
My second script named (test2.py):
def mult():
x= 2 * 4
print(x)
return x
def addx():
sum = x + 2
print("sum",sum)
Error:
NameError: name 'x' is not defined
In the function addx() you haven't declared x. I believe you want x from mult. So you can do something like this
def addx():
x = mult()
sum = x + 2
print("sum",sum)
You should use the return value of mult, to pass it to your second function addx as a parameter.
def todo (self):
x = mult()
addx(x)
I advise you to read the Python doc section about function : https://docs.python.org/fr/3/tutorial/controlflow.html#defining-functions
In test2.py, you have not defined x
def addx():
sum = x + 2
print("sum",sum)
The problem above is that the computer doesn't know what x is. You could pass it as a parameter:
def addx(x):
sum = x + 2
print("sum", sum)
and change your code to:
from test2 import *
class Test():
def todo(self):
addx(x=mult()) # whatever number you want
if __name__ == '__main__':
Test().todo()
def move(self):
z = self.comboBox.currentText()
print(z)
Hospital = newtest.my_function()
i = Hospital.index(z)
print('The index of :', i)
user = newuser.my_function()
global (user[i])
print (user[i])
return user[i]
def my_doc():
url = 'https://test.com/steth/get-list'
myobj = {'mongoId': 'user[i]'}
x = requests.post(url, data = myobj)
y=x.json();
print(y)
my_doc()
I need to get user[i] in the second function my_doc.so i made user[i] global.But it is showing syntax error as
global (user[i])
^
SyntaxError: invalid syn
Just get the return of one function, and input it into the other.
class Obj:
def func1(self):
return "something"
# Assuming func2 is inside class.
def func2_class(self, something):
print(something)
# Assuming func2 is outside class.
def func2(something):
print(something)
obj = Obj()
something = obj.func1()
func2(something) # Outside class.
obj.func2(something) # Inside class.
So here's an extension to this question: https://stackoverflow.com/a/37568895/2290820
on how to optionally Enable or Disable Decorator on a Function.
On those lines, I came up with something like this to make decorator get invoked on a recursive call:
def deco(f):
def fattr(attr):
f.attr = attr
def closure(*args):
f(*args)
f.unwrap = f
f.closure = closure
return f
return fattr
#deco
def printa(x):
if x > 1:
print x
return printa(x-1)
else:
print x
return
printa({1:1})(5)
# do the same call w/o deocorator
def finta(x):
if x > 1:
print x
return finta(x-1)
else:
print x
return
finta(5) # this works
to experiment with decorators on a recursive function. Clearly, printa recursive version is not behaving the way it should be.
I could do
g = printa({1:1})
g.closure(5)
to turn on the decorator option or not use that option. Anyway, regardless of good or bad design, How can I make decorator get invoked on a recursive call?
In your deco you have an assignment f.attr = attr that "eats" your argument after first recursive call. Your should modify your recursive call this way:
def deco(f):
def fattr(attr):
f.attr = attr
def closure(*args):
f(*args)
f.unwrap = f
f.closure = closure
return f
return fattr
#deco
def printa(x):
if x > 1:
print x
return printa(None)(x-1) # None will be assigned to f.attr
else:
print x
return
printa({1:1})(5)
5
4
3
2
1