Find values in a shuffled list of subscriptions by one of the values
I have a list if lists
items = [
["e",None,None],
["pork","pork.png","meat"],
["beef","b.png","meat"],
["cheese","c.png","not"],
]
items_list = ["e","beef","pork","beef"]
shuffle(items_list)
how can I print the second or third value in the sublists without their index?
for i in items_list:
print ???
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You can store the signatures items
in a dictionary indexed by their first elements.
from random import shuffle
items = [
["e",None,None],
["pork","pork.png","meat"],
["beef","b.png","meat"],
["cheese","c.png","not"],
]
items_dict = {u[0]: u for u in items}
items_list = ["e","beef","pork","beef"]
shuffle(items_list)
for s in items_list:
print(s, items_dict[s])
Output
beef ['beef', 'b.png', 'meat']
e ['e', None, None]
beef ['beef', 'b.png', 'meat']
pork ['pork', 'pork.png', 'meat']
To print the second element (e.g. PNG):
for s in items_list:
print(s, items_dict[s][1])
Output
e None
beef b.png
beef b.png
pork pork.png
This is quite efficient because the new lists are not executed: the lists in items_dict
are the same list objects that are in items
. Therefore, if you want, you can mutate these lists either through items
or items_dict
.
items_dict["cheese"].append("cheddar")
print(items[3])
items[0][2] = "something"
print(items_dict["e"])
Output
['cheese', 'c.png', 'not', 'cheddar'] ['e', None, 'something']
You don't need items_dict
, but the alternative is a double loop for
, which becomes very inefficient if items
large.
for s in items_list:
for seq in items:
if seq[0] == s:
print(s, seq)
break
Output
beef ['beef', 'b.png', 'meat']
e ['e', None, None]
beef ['beef', 'b.png', 'meat']
pork ['pork', 'pork.png', 'meat']
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How about this:
from random import shuffle
items = [
["e",None,None],
["pork","pork.png","meat"],
["beef","b.png","meat"],
["cheese","c.png","not"],
]
items_list = ["e","beef","pork","beef"]
shuffle(items_list)
for item in items_list:
for orig_item in items:
try:
orig_item.index(item)
print(item, orig_item)
except ValueError:
pass
Output:
beef ['beef', 'b.png', 'meat']
pork ['pork', 'pork.png', 'meat']
e ['e', None, None]
beef ['beef', 'b.png', 'meat']
It might be slower than @ PM 2Ring's answer (I haven't measured) due to two loops and exception handling.
use orig_item[1]
to access the second item of the sublist
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