I want to do this:
[("item1", "item2"), ("item3", "item4")]
# ->
[("item1", ("item2", True)), ("item3", ("item4", True))]
How can I do this? I have a list of tuples, and the second element in each tuple needs to be it's own tuple, and I need to append an element to the sub tuple.
an option is to use a list-comprehension:
lst = [("item1", "item2"), ("item3", "item4")]
res = [(a, (b, True)) for a, b in lst]
print(res) # [("item1", ("item2", True)), ("item3", ("item4", True))]
You can use list comprehension to complete this:
material = [("item1", "item2"), ("item3", "item4")]
# ->
expected = [("item1", ("item2", True)), ("item3", ("item4", True))]
actual = [(elt[0], (elt[1], True)) for elt in material]
assert actual == expected
Easy answer
lst = [("item1", "item2"), ("item3", "item4")]
new_lst = []
for f,s in lst:
new_lst.append((f, (s, True)))
print(new_lst)
OUTPUT
[("item1", ("item2", True)), ("item3", ("item4", True))]
USING LIST COMPREHENSION
lst = [("item1", "item2"), ("item3", "item4")]
new_lst = [(f, (s, True)) for f,s in lst]
print(new_lst)
OUTPUT
[("item1", ("item2", True)), ("item3", ("item4", True))]
Another way to accomplish this would be to use the map function.
This is what you could do:
def map_func(item):
return (item[0], (item[1], True))
array = [("item1", "item2"), ("item3", "item4")]
new_array = list(map(map_func, array))
print(new_array) # [('item1', ('item2', True)), ('item3', ('item4', True))]
You could use lambda too (in order to shorten the code), like so:
new_array = list(map(lambda item: (item[0], (item[1], True)), array))
