You haven't created three different empty lists. You've created one empty list, and then created a new list with three references to that same empty list. To fix the problem use this code instead:
listy = [[] for i in range(3)]
Running your example code now gives the result you probably expected:
>>> listy = [[] for i in range(3)]
>>> listy[1] = [1,2]
>>> listy
[[], [1, 2], []]
>>> listy[1].append(3)
>>> listy
[[], [1, 2, 3], []]
>>> listy[2].append(1)
>>> listy
[[], [1, 2, 3], [1]]
You haven't created three different empty lists. You've created one empty list, and then created a new list with three references to that same empty list. To fix the problem use this code instead:
listy = [[] for i in range(3)]
Running your example code now gives the result you probably expected:
>>> listy = [[] for i in range(3)]
>>> listy[1] = [1,2]
>>> listy
[[], [1, 2], []]
>>> listy[1].append(3)
>>> listy
[[], [1, 2, 3], []]
>>> listy[2].append(1)
>>> listy
[[], [1, 2, 3], [1]]
[[]]*3 is not the same as [[], [], []].
It's as if you'd said
a = []
listy = [a, a, a]
In other words, all three list references refer to the same list instance.
Videos
say i have array [[1,2,3,4,5]], how can i duplicate it 5 times to [[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5],[1,2,3,4,5]] ?
You do not "declare" arrays or anything else in python. You simply assign to a (new) variable. If you want a multidimensional array, simply add a new array as an array element.
arr = []
arr.append([])
arr[0].append('aa1')
arr[0].append('aa2')
or
arr = []
arr.append(['aa1', 'aa2'])
There aren't multidimensional arrays as such in Python, what you have is a list containing other lists.
>>> arr = [[]]
>>> len(arr)
1
What you have done is declare a list containing a single list. So arr[0] contains a list but arr[1] is not defined.
You can define a list containing two lists as follows:
arr = [[],[]]
Or to define a longer list you could use:
>>> arr = [[] for _ in range(5)]
>>> arr
[[], [], [], [], []]
What you shouldn't do is this:
arr = [[]] * 3
As this puts the same list in all three places in the container list:
>>> arr[0].append('test')
>>> arr
[['test'], ['test'], ['test']]
You can iterate through the items of the lists and then add them to your masterlist:
a = [[1,2,3],
[1,2,3]]
b = [[4,5,6],
[4,5,6]]
masterlist = []
for aa,bb in zip(a,b): # loop over lists
for itema, itemb in zip(aa,bb): # loop over items in list
masterlist = masterlist + [[itema, itemb]]
output:
[[1, 4], [2, 5], [3, 6], [1, 4], [2, 5], [3, 6]]
If you use numpy,this is really easy
import numpy as np
a = np.array([[1,2,3],
[1,2,3]])
b = np.array([[4,5,6],
[4,5,6]])
fl = np.vstack(np.dstack((a,b)))
output
array([[1, 4],
[2, 5],
[3, 6],
[1, 4],
[2, 5],
[3, 6]])
I'm new to python and I've been given a task that requires me to append/copy a 1d list multiple times to form a 2d list, while also modifying the values of the 2d list.
But every time I've attempted it I get the same issue, that whenever I change the value of one entry in the 2d list, it changes that value for the entire column which I'm doing by "2dlist[x][y] = 1".
I've tried the following.
2dlist.Append(1dlist)
2dlist += 1dlist
2dlist = [1dlist for i in range(number)]
1dlist *= number
And several other methods that all resulted in the same issue or worse.
I'm not allowed to import libraries and I've ran out ideas. Am I mis-understanding something about python lists? Is "2dlist[x][y] = 1" at fault? Any and all help/tips/resources is GREATLY appreciated.
Example
1dlist = [1, 2, 3, 4, 5]
and I'm trying to get 2dlist to equal
[[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5],
[1, 2, 3, 4, 5]]
Then modifying each value to get
[[1, 0, 3, 0, 5], #multiply 1, remove even
[0, 4, 0, 8, 0], #multiply 2, remove odd
[3, 0, 9, 0, 15], #multiply 3, remove even
[0, 8, 0, 16, 0]] #multiply 4, remove odd
What I'm getting
[[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0]]
Because it removes the evens in all rows then all odds in each column