How can I create as many empty lists as a for loop variable determines ?
As in:
for i in range(3):
list_{i} = [] #this is just an idea
In the end of the for loop I'd have:
list_0 = [] #empty, I could add elements later.
list_1 = [] #empty
list_2 = [] #empty
I'm new to Python so please if this question is a duplicate, let me know but I haven't found any question with the same needs.
lists_= [[] for i in range(3)]
print(lists_)
OUTPUT:
[[], [], []]
and then if you want to add elements to a specific list_:
lists_[0].append(1)
lists_[0].append(2)
lists_[0].append(3)
print(lists_[0])
OUTPUT:
[1, 2, 3]
EDIT:
Since OP mentioned that range is conditional:
cond_1 = False
if cond_1:
x = 3
else:
x = 2
lists_= [[] for i in range(x)]
print(lists_)
OUTPUT:
[[], []]
[1, 2, 3]
ifcondition, so I can't tell what's the range.
if array[i][0] == 1 than create list, than i is incremented, what means that iis checking all the 100 rows. In the end if I have 20 rows with the first element equal to 1 (array[i][0] == 1), I'd need 20 lists.Try this:
for i in range(3):
globals()[f'list_{i}'] = [] # Only Python3.6 or higher
for python lower than 3.6:
for i in range(3):
globals()['list_{}'.format(i)] = []
then you will have list_0, list_1 and list_2 equals to [].
Either via a list comprehension:
lst = [[] for i in range(3)]
Or via .append():
lst = list()
for i in range(3):
lst.append([])
print(lst)
1. You can create a list of a lists the next way using for loop
>>> pool = []
>>> for i in range(3):
>>> pool.append([])
2. You can create it dynmicly
>>> pool = [[] for i in range(3)]
3. However if you want specify names, maby it is better to store them in a dict
>>> pool = dict([(i, []) for i in range(3)])
The output will be:
>>> pool
{0: [], 1: [], 2: []}
4. The same example with the readable names
>>> pool = dict([(i, []) for i in ['list1', 'list2', 'list3']])
The output in that example will be:
>>> pool
{'list1': [], 'list3': [], 'list2': []}
collections.defaultdict. There's a high chance this is the solution you need.