I am new to programming and trying to understand the syntax.
in my example, i would like to have a result of: table = [[1,2,3], [2,4,6], [3,6,9]] but somehow i get and empty list. I am not sure where to fix it.
row = [1, 2, 3]
col = [1, 2, 3]
li = []
table = []
for x in row:
for n in col:
li.append(x*n)
table.append(li)
li.clear()
print(table)
Under the hood, lists are pointers. This means that when you append a list to another list, it is not creating a new list, it is using the same list.
So whenever you do li.clear, it gets rid of the internal list elements...
you're gonna want this instead...
row = [1, 2, 3]
col = [1, 2, 3]
table = []
for x in row:
li = []
for n in col:
li.append(x*n)
table.append(li)
print(table)
Create a new list each time, so it's a separate object in memory for each append operation
using your own code:
row = [1, 2, 3]
col = [1, 2, 3]
table = []
for x in row:
li = []
for n in col:
li.append(x*n)
table.append(li)
print(table)
you will then get:
row = [1,2,3]
col = [1,2,3]
table = []
for x in row:
li = []
for n in col:
li.append(x*n)
table.append(li)
The difference is making the li = [] variable inside the first level of your for loop. This essentially "clears" it out each time.
You need to look into what is mutabe and immutable in python
Let me explain in brief what happens in your code
row = [1, 2, 3]
col = [1, 2, 3]
li = []
table = []
for x in row:
for n in col:
li.append(x*n)
table.append(li) # everything is good until here
li.clear() # but here you are sayning that li is empty so it also goes and changes table.append(li) to table.append(li.clear()), because li is a list and lists are mutable objects.
# what you can do is create another new object (which happens to be named li: li=[] so the table.append(li) will not change to table.append([])
change your code to
row = [1, 2, 3]
col = [1, 2, 3]
li = []
table = []
for x in row:
for n in col:
li.append(x*n)
table.append(li)
li = []
print(table)
li.clear()?table.append(li[:])- if you clear the same list that you're appending to the table, obviously you'll end up with[]. Append a copy instead[[1,2,3],[1,2,3,2,4,6],[1,2,3,2,4,6,3,6,9]]