how can we take input in lambda function?
z = lambda a,b,c,d : a+(b-c)*d
y = z(input("enter numbers: "))
print(y)
I expect input of the above expressions.
5 Answers
input returns a string, you need to:
splitit- cast each part to int using
map - unpack it to different args using
*unpacking
try this:
z = lambda a, b, c, d: a + (b - c) * d
y = z(*map(int,input("enter numbers: ").split()))
print(y)
example run:
enter numbers: 1 2 3 4
-3
2 Comments
Kenny Ostrom
The important point is there was nothing special about lambda. It's just that you have to pass the right arguments to call a function (defined or lambda).
Adam.Er8
@KennyOstrom yes, good clarification,
lambda is just a shorthand way of defining a single-expression function.A simple example for one variable:
In [1]: z = lambda x : x+1
In [2]: z(int(input()))
4
Out[2]: 5
In [3]:
Comments
That's neither readable nor flexible, but you can always embed a call to input in the lambda:
z = lambda: int(input('a:')) + int(input('b:'))
1 Comment
chepner
This doesn't allow you to recover from the
ValueError that int will raise if the input isn't valid.input returns a string, so you'll need to convert that first:
x = input('enter numbers: ') # 1 2 3 4
'1 2 3 4'
# use split
x = input('enter numbers: ')
x = [int(num) for num in x.split()]
# then unpack with the * syntax
c = lambda a, b, c, d: a+(b-c)*d
c(*x)
-3
Comments
As rightly pointed out by Adam, the input function would get you a string. You can split the string and get the comma separated values and then convert them to integer like below
z = lambda l : int(l[0])+(int(l[1])-int(l[2]))*int(l[3])
y = z(input("enter numbers: ").split(','))
print(y)
An alternate way to achieve this would be:
y = lambda stringFromTerminal : [int(eachString) for eachString in stringFromTerminal.split(',')]
x = y(input("enter numbers: "))
print(x[0]+(x[1]-x[2])*x[3])
lambdais usually reserved for key functions and inline definitionsdefstatement.