I would like to get the return value from this function and assign it a variable if I put in "5" then we get 15 out but I want to keep using the updated variable afterwards in the next function. I would like to avoid joining the functions too.
To clarify, if you input this function then the result is 5+10 and 5+47. What I want as an answer is 5+10 and then 15+47.
y = input("here you input your number: ")
def this_is_function(x):
return int(x) + 10
def this_is_function2(x):
return int(x) + 47
this_is_function(y)
this_is_function2(y)
Currently you are ignoring the result of this_is_function
(which is 15) and are just passing the same value y
(which is 5) again to this_is_function2
.
Save the result of this_is_function
in a variable instead, and pass it to this_is_function2
:
z = this_is_function(y) # z is 5 + 10 = 15 (y is still 5)
result = this_is_function2(z) # result is 15 + 47
If you don't need the original y
anymore, you can also reuse the y
variable name instead of using a new name, z
, for the intermediate result.
y = this_is_function(y) # y is now 5 + 10 = 15 (no longer 5)
result = this_is_function2(y) # result is 15 + 47
If you are wanting the input to the second function (equation) to be the output of the first, then you can do just that:
y = input("here you input your number: ")
def this_is_function(x):
return int(x) + 10
def this_is_function2(x):
return int(x) + 47
^leaving all this the same
Then using this:
this_is_function2(this_is_function(y))
You can now see that the input to this_is_function2
is this_is_function(y)
and the input to this_is_function
is y
So writing it as I would solve a math problem - and as python performs this when debugging:
this_is_funtion2(this_is_function(y)) # Original line
this_is_funtion2(this_is_function(5)) # y = 5
this_is_funtion2(15) # this_is_function(5) = 15
62 # this_is_function2(15) = 62
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.