I want to add an item to my list only if a variable is true, but I want to do that in one line to not make my program longer and crappy (because I want to do that on a lot of variables). Let's say I want to return a list, but include items in it only if x and etc are true.
I want to use something that won't add any data to the list, because None is considered as an item.
# Note: The variables name may change, so please make it compatible.
return [data1 if x else None, data2 if y else None]
<< x = False, y = False = [None, None], I want []
Example
# User choose if he wants data(x) by changing the default value of want_data(x)
def func(want_data1=True, want_data2=True):
return [data1 if want_data1 else None, data2 if want_data2 else None]
print(func(False, False))
<< [None, None] # Not what I want! Excepted output is []
You are working with ternary expressions and they only allow the following syntax, which is equivalent to (expr1, expr2)[condition]
expr1 if cond else expr2
Even if you try python short-circuiting, it will still return False if x is false and data1 when x is True.
[x and data1, y and data2]
For your case, you will have to do a list comprehension on top of the ternary opertion you have created as -
data1 = 'hi'
data2 = 'there'
x, y = True, False
[i for i in [data1 if x else None, data2 if y else None] if i]
['hi']
Your function with minimal modification with the above approach:
def func(want_data1=True, want_data2=True):
l = [data1 if want_data1 else None, data2 if want_data2 else None]
return [i for i in l if i]
Alternate approach - If you are open to numpy, its super easy to mask the list then try this:
import numpy as np
def func(want_data1=True, want_data2=True):
return np.array([data1, data2])[[want_data1,want_data2]].tolist()
With a dictionary -
import numpy as np
def func(want_data1=True, want_data2=True):
d = {'data1': data1, 'data2': data2}
return dict(np.array(list(d.items()))[[want_data1,want_data2]])