Hope you can help me with this:
previous=[[0, 0, 1, 1], [0, 0, 0, 0], [0, 0, 1, 2], [0, 0, 1, 2], [0, 0, 1, 1], [0, 1, 1, 2], [0, 1, 1, 1], [0, 0, 1, 2], [0, 1, 1, 2], [0, 0, 0, 0]]
type(previous)
Out[67]: list
previous[0][-1]
Out[66]: 1
previous[1][-1]
Out[65]: 0
for p in previous:
print(p)
[0, 0, 1, 1]
[0, 0, 0, 0]
[0, 0, 1, 2]
[0, 0, 1, 2]
[0, 0, 1, 1]
[0, 1, 1, 2]
[0, 1, 1, 1]
[0, 0, 1, 2]
[0, 1, 1, 2]
[0, 0, 0, 0]
My Question, I need to get the 3 value from each element in the sub list and append it to create another list.
The result should be:
[1,0,1,1,1,1,1,1,1,0]
I tried this but without success:
mylist=[]
for p in previous:
for x in p:
mylist.append(p[x][-1])
print(mylist)
and:
for p in previous:
for x in p:
print ([p[x]])
p is the sublist, you don't need to loop over it again with for x in p: to get the third element (so your code actually raises an IndexError when trying to index the sublist p if x is out of its bound, and will raise an TypeError when trying to use p[x][-1] to index the number at p[x]), still, p[-1] is the last element (the fourth in your case), you should either use positive indexing:
p[2] # the third element
Or use negative indexing to get the next-to-last element:
p[-2]
But if the lists should grow any longer, positive indexing is better since it won't be affected by elements added to the end of the list.
You could also write a list comprehension instead of a for loop + append:
mylist = [p[2] for p in previous]
print(mylist)
Output:
[1, 0, 1, 1, 1, 1, 1, 1, 1, 0]
In your loop p is the sublist. The first time through the loop p is
[0, 0, 1, 1]
Python does zero based indexing (the first element is always at index 0). So p[0] is the first element. The third element would be p[2].
If you use negative indexes, -1 gives the last element and -2 is the next-to-last or penultimate element.
To modify your solution to work, you can remove the inner loop. You don't need it. You already have the sublist.
mylist=[]
for p in previous:
mylist.append(p[2])
print(mylist)
Related
So I am trying to create an NxN 2D array and then change its diagonal elemets to 1. Here is my code:
arr=[1,1,1,2,2,2]
table=[[0]*len(arr)]*len(arr)
for i in range(0,len(arr)):
table[i][i]=1
print(table)
However, whenever I run this code, I get this output:
[[1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1],
[1, 1, 1, 1, 1, 1]]
I am looking to get this:
[[1, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 1]]
I have been staring at my code for hours and I cannot figure out what's wrong
The interesting thing about this is that you are really only editing one list in the for loop, but there are just five pointers to that list. (In this case, the list would be [0, 0, 0, 0, 0, 0].) You can see this by printing the id of each list in table by using id():
>>> for t in table:
print(id(t))
2236544254464
2236544254464
2236544254464
2236544254464
2236544254464
2236544254464
Your numbers are likely different than mine, but they are all the same number, nevertheless. You also can see that the edits to one list are applied to the others in table by putting a print(table) statement after each index assignment statement.
So in order to 'fix' this, I would recommend using list comprehension instead. For example:
table = [[0]*len(arr) for _ in range(len(arr))]
If you checkout the ids of each list:
>>> for t in table:
print(id(t))
2236544617664
2236544616064
2236544616320
2236544615872
2236544618368
2236544622720
Since they are different, you can now use the method for changing only the diagonals:
>>> for i in range(0,len(arr)):
table[i][i]=1
>>> table
[[1, 0, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0],
[0, 0, 0, 1, 0, 0],
[0, 0, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 1]]
Your 2D "array" contains 6 lists which are the same list. Changes to any of those lists will also be reflected in the other lists. Consider this:
>>> l = [0] * 6
>>> x = [l]
>>> l[0] = 1
>>> l
[1, 0, 0, 0, 0, 0]
>>> x
[[1, 0, 0, 0, 0, 0]]
>>> x = [l, l, l]
>>> x
[[1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0]]
>>> x[-1][-1] = 100
>>> x
[[1, 0, 0, 0, 0, 100], [1, 0, 0, 0, 0, 100], [1, 0, 0, 0, 0, 100]]
This is because the list x contains the list l, so any changes to l are also seen through the reference to the same list in x.
The problem is when multiplying mutable objects because it creates multiple references to the same mutable object.
You should initialise your table like this:
table = [[0 for j in range(len(arr))] for i in range(len(arr))]
or
table = [[0] * len(arr) for i in range(len(arr))]
which, despite the use of multiplication, works because each list is different.
You can create your table and populate it simultaneously in nested loops:
arr=[1,1,1,2,2,2]
table = []
for i in range(len(arr)):
table.append([0]*len(arr))
for j in range(len(arr)):
if i == j:
table[i][j] = 1
print(table)
#[[1, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 1, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 0, 1]]
Interesting.
Try to use numpy to avoid list trap:
import numpy as np
org_row = [0]*5
l = [org_row]*5
x = np.array(l, np.int32)
for i in range(len(x)):
x[i][i]=1
print(x)
output>:
output>
[[1 0 0 0 0]
[0 1 0 0 0]
[0 0 1 0 0]
[0 0 0 1 0]
[0 0 0 0 1]]
Expected Output:
indenitiy_matrix(3)
[[1, 0, 0], [0, 1, 0], [0, 0, 1]]
Actual Output with Error:
indenitiy_matrix(3)
[[1, 1, 1], [1, 1, 1], [1, 1, 1]]
def identity_matrix(n):
list_template = [[]]
list_n = list_template*n
for sub_l in list_n:
sub_l.append(0)
for val in range(n):
# I have the feeling that the problem lies somewhere around here.
list_n[val][val]=1
return(list_n)
list_template*n does not create n copies, instead but all those n copies reference to only one copy. For example see this
a = [[0,0,0]]*2
# Now, lets change first element of the first sublist in `a`.
a[0][0] = 1
print (a)
# but since both the 2 sublists refer to same, both of them will be changed.
Output:
[[1, 0, 0], [1, 0, 0]]
Fix for your code
def identity_matrix(n):
list_n = [[0]*n for i in range(n)]
for val in range(n):
list_n[val][val]=1
return list_n
print (identity_matrix(5))
Output:
[[1, 0, 0, 0, 0],
[0, 1, 0, 0, 0],
[0, 0, 1, 0, 0],
[0, 0, 0, 1, 0],
[0, 0, 0, 0, 1]]
No, the problem lies here:
list_template = [[]]
list_n = list_template*n
After this, try doing:
list_n[0].append(1) # let's change the first element
The result:
[[1], [1], [1], [1], [1]]
is probably not what you expect.
Briefly, the problem is that after its construction, your list consists of multiple references to same list. A detailed explanation is at the link given by #saint-jaeger : List of lists changes reflected across sublists unexpectedly
Finally, the numpy library is your friend for creating identity matrices and other N-dimensional arrays.
gff_lijst_gefilterd = []
for x in range(1, len(gff_lijst)):
if filter_genID[x] == x:
gff_lijst_gefilterd.append(gff_lijst_gefilterd[x])
return gff_lijst_gefilterd
So i want this code to loop over gff_lijst which is a 2d list. It checks if the keys of the dictionary are the same of the 8th value of the list and if it does than it appends the value of the list to
the gff_lijst_gefilterd
I hope you can help me!
You're much better off just iterating through your 2D list and then checking if the last element is in your dict.
Here is an example:
# this is just a test list
gff_lijst = [[1, 0, 1, 0, 1, 0, 1, 5],
[1, 0, 1, 0, 1, 0, 1, 4],
[1, 0, 1, 0, 1, 0, 1, 3],
[1, 0, 1, 0, 1, 0, 1, 8],
[1, 0, 1, 0, 1, 0, 1, 9],
[1, 0, 1, 2, 1, 5, 1, 4],
[1, 0, 1, 0, 1, 0, 1, 6]]
# and a test dict
filter_genID = {3 : 'test', 4 : 'test2'}
gff_lijst_gefilterd = []
# iterate through your 2d list
# in this case i will be each list in your 2d list
for i in gff_lijst:
# check if the last element of the current list is in your dict keys
if i[-1] in filter_genID:
# if it is we add it to our list
gff_lijst_gefilterd.append(i)
print(gff_lijst_gefilterd)
What is the simplest and reasonably efficient way to slice a list into a list of the sliced sub-list sections in a reverse manner?
Here is the portion of my code that groups list into sublist:
binary1 = [1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1]
process1 = [binary1[i:i+4] for i in range(0, len(binary1), 4)]
print(process1)
Result: [[1, 0, 0, 1], [1, 0, 1, 0], [1, 0, 1, 1], [0, 1]]
However the result above is really not what I want is it will group in a reversal way, here is the result that I expected/want:
Result: [[1, 0], [0, 1, 1, 0], [1 0, 1, 0], [1, 1, 0, 1]]
I hope you could help me. Thank you!
binary1 = [1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1]
rest = len(binary1) // 4
print([binary1[:rest-1]] + [binary1[i:i+4] for i in range(rest-1, len(binary1), 4)])
Will print:
[[1, 0], [0, 1, 1, 0], [1, 0, 1, 0], [1, 1, 0, 1]]
I'm trying to construct permutations of a list with the i-th bit flipped every time.
For example:
With input:
[1,0,0,0]
To get:
[0,0,0,0]
[1,1,0,0]
[1,0,1,0]
[1,0,0,1]
I wrote a method which given a list returns the same list with the bit at position p changed:
def flipBit(l,p):
l[p] = ~l[p]&1
return l
And I'm trying to apply it using a map(), but I can't even get a basic example working:
p=list(map(flipBit, [[1,0,0]]*3,range(3))))
This is what it returns:
[[0, 1, 1], [0, 1, 1], [0, 1, 1]]
while expecting:
[[0, 0, 0], [1, 1, 0], [1, 0, 1]]
What I'm I doing wrong? (if anyone can suggest an even shorter code for this maybe without using a flipbit method I'd appreciate it as I won't really use flipbit other than this purpose and I want to keep the code concise and clean)
The issue is that [[1,0,0]]*3 creates a list containing three references to the same sublist. When you change one sublist, they all change.
Here is one way to fix this:
>>> list(map(flipBit, [[1,0,0] for _ in range(3)], range(3)))
↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑
[[0, 0, 0], [1, 1, 0], [1, 0, 1]]
And here is a way to implement this functionality without using a helper function:
>>> l = [1, 0, 0]
>>> [l[:i] + [1-l[i]] + l[i+1:] for i in range(len(l))]
[[0, 0, 0], [1, 1, 0], [1, 0, 1]]
The code you posted is was not valid at all, but I presume you need this:
>>> p=list(map(lambda x: flipBit([1, 0, 0], x), range(3)))
>>> p
[[0, 0, 0], [1, 1, 0], [1, 0, 1]]
Basically, you map with a lambda function, that partially applies [1, 0, 0] as l, and then takes each element in range(3) and applies it to p.