I have a list of lists that correspond to lines in file, with multiple columns.
[[col1, col2, col3], [elem1, elem2, elem3], [elem4, elem5, elem6]]
I want to check if (for example) elem3 is in any of the lists, and if it is, go into that list. (really I have a list of things I need to check, so it's a list that probably contains elem3, elem5, elem7.... etc)
the shortest way is to use list comprehensions and list comprehensions sometimes are more faster than simple for loop
your list:
list1 = [ ["col1", "col2", "col3"], ["elem1", "elem2", "elem3"], ["elem4", "elem5", "elem6"] ]
your element to find:
to_find = "col1"
your function to "go into that list":
def do_something(sub_list):
print (sub_list)
and the list comprehension that will find your element and call function with list that have it:
[do_something(sub_list) for sub_list in list1 if to_find in sub_list]
You can do something like this:
def in_list(list_of_lists, item):
for list_ in list_of_lists:
if item in list_:
return list_
EDIT:
Here is a recursive version for the heck of it:
def in_list(list_of_lists, item):
if not list_of_lists:
return None
if item in list_of_lists[0]:
return list_of_lists[0]
return in_list(list_of_lists[1:], item)
in doesn't iterate over the list, so you're just going to iterate over the list_of_lists. It should not be slow at all.
in doesn't iterate over the list" Yes it does. item in lst is a O(n) operation that iterates over lst until item is found, or lst is exhausted.item in one of the sub-lists, you wouldn't want a list comprehension here, because you can't short-circuit it as soon as you find a match. With this answer, you stop as soon as item is found.I came across a similar problem. I had a list of tuples, with tile numbers:
data = [(0, 0, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 0),
(0, 23, 24, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 0),
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21)]
I needed to find out where in the list of tuples a particular tile number is located, so:
to_find = 42
for i in range(len(data)):
if to_find in data[i]:
print (i, data[i].index(to_find))
break
You can check if a value exists in a list of lists as shown below:
list_of_lists = [['John', 'David'], ['Anna', 'Lisa']]
print(any('Lisa' in list for list in list_of_lists)) # True
print(any('Tom' in list for list in list_of_lists)) # False
