Things get wrong when appending a list to a list in python
I am programming a simple software for practicing.
def check(num, lst):
for i in lst:
if num == i:
return False
else:
lst.append(num)
return True
timelst =
for i1 in range(1,7):
usednum = [i1]
for i2 in range(1,7):
if not check(i2, usednum):
continue
for i3 in range(1,7):
if not check(i3, usednum):
continue
for i4 in range(1,7):
if not check(i4, usednum):
continue
for i5 in range(1,7):
if not check(i5, usednum):
continue
for i6 in range(1,7):
if not check(i6, usednum):
continue
else:
print(usednum) #print the appending list before actual appending
timelst.append(usednum)
usednum.pop()
break
usednum.pop()
When running this, this is what I want the timelst to be like:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 6, 5],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 6, 5],
...
]
However this is what I actually get:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
...
]
I am really confused about this and got stuck for a long time. I tried printing the 'usednum' list before appending it, and it returns perfectly what I want.
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 6, 5]
[2, 1, 3, 4, 5, 6]
[2, 1, 3, 4, 6, 5]
#printed lists while running the program
But the same problem still occurs everytime when I check the timelst after executing the program.
I used python 3.7 with spyder. There should not be a problem with my compiler, as I have tried running this on ipython and I still got the same result.
Can anyone please help me solving this problem? Thanks!
python list
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
add a comment |
I am programming a simple software for practicing.
def check(num, lst):
for i in lst:
if num == i:
return False
else:
lst.append(num)
return True
timelst =
for i1 in range(1,7):
usednum = [i1]
for i2 in range(1,7):
if not check(i2, usednum):
continue
for i3 in range(1,7):
if not check(i3, usednum):
continue
for i4 in range(1,7):
if not check(i4, usednum):
continue
for i5 in range(1,7):
if not check(i5, usednum):
continue
for i6 in range(1,7):
if not check(i6, usednum):
continue
else:
print(usednum) #print the appending list before actual appending
timelst.append(usednum)
usednum.pop()
break
usednum.pop()
When running this, this is what I want the timelst to be like:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 6, 5],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 6, 5],
...
]
However this is what I actually get:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
...
]
I am really confused about this and got stuck for a long time. I tried printing the 'usednum' list before appending it, and it returns perfectly what I want.
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 6, 5]
[2, 1, 3, 4, 5, 6]
[2, 1, 3, 4, 6, 5]
#printed lists while running the program
But the same problem still occurs everytime when I check the timelst after executing the program.
I used python 3.7 with spyder. There should not be a problem with my compiler, as I have tried running this on ipython and I still got the same result.
Can anyone please help me solving this problem? Thanks!
python list
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
1
Multiple nested loops like this is generally considered bad programming practice, especially when the loops do the same thing. It's not entirely clear to me what you want the output to be ... what is the intention of this code?
– Jonah Bishop
Nov 19 at 22:04
Basically I want to append every possible combinations of the numbers under six to a list, such as 123456 123465 123546 123564 ... 654312 654321, and this is the most efficient way I can think to achieve this, rather than append all possible lists and remove the ones with repeating numbers. The program is not totally finished, it should work, ideally and maybe logically, if I add timelst.pop() at the end of every loop after the previous loop finishes. I did try it, but it gave me a much worse result like [[1], [1], [1], ...., [6], [6], [6]]. I realise maybe they are caused by the same problem.
– PotatoPoweredLaptop
Nov 19 at 22:30
check the indentation of yourcheckfunction.
– Paul H
Nov 19 at 22:56
Sorry that was just a silly mistake when I was transferring my code to my post. It was correct in my compiler and that did not cause any problem, and the indentation of function has been corrected in my post
– PotatoPoweredLaptop
Nov 19 at 23:20
add a comment |
I am programming a simple software for practicing.
def check(num, lst):
for i in lst:
if num == i:
return False
else:
lst.append(num)
return True
timelst =
for i1 in range(1,7):
usednum = [i1]
for i2 in range(1,7):
if not check(i2, usednum):
continue
for i3 in range(1,7):
if not check(i3, usednum):
continue
for i4 in range(1,7):
if not check(i4, usednum):
continue
for i5 in range(1,7):
if not check(i5, usednum):
continue
for i6 in range(1,7):
if not check(i6, usednum):
continue
else:
print(usednum) #print the appending list before actual appending
timelst.append(usednum)
usednum.pop()
break
usednum.pop()
When running this, this is what I want the timelst to be like:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 6, 5],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 6, 5],
...
]
However this is what I actually get:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
...
]
I am really confused about this and got stuck for a long time. I tried printing the 'usednum' list before appending it, and it returns perfectly what I want.
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 6, 5]
[2, 1, 3, 4, 5, 6]
[2, 1, 3, 4, 6, 5]
#printed lists while running the program
But the same problem still occurs everytime when I check the timelst after executing the program.
I used python 3.7 with spyder. There should not be a problem with my compiler, as I have tried running this on ipython and I still got the same result.
Can anyone please help me solving this problem? Thanks!
python list
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
I am programming a simple software for practicing.
def check(num, lst):
for i in lst:
if num == i:
return False
else:
lst.append(num)
return True
timelst =
for i1 in range(1,7):
usednum = [i1]
for i2 in range(1,7):
if not check(i2, usednum):
continue
for i3 in range(1,7):
if not check(i3, usednum):
continue
for i4 in range(1,7):
if not check(i4, usednum):
continue
for i5 in range(1,7):
if not check(i5, usednum):
continue
for i6 in range(1,7):
if not check(i6, usednum):
continue
else:
print(usednum) #print the appending list before actual appending
timelst.append(usednum)
usednum.pop()
break
usednum.pop()
When running this, this is what I want the timelst to be like:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 6, 5],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 6, 5],
...
]
However this is what I actually get:
[[1, 2, 3, 4, 5, 6],
[1, 2, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
[2, 1, 3, 4, 5, 6],
...
]
I am really confused about this and got stuck for a long time. I tried printing the 'usednum' list before appending it, and it returns perfectly what I want.
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 6, 5]
[2, 1, 3, 4, 5, 6]
[2, 1, 3, 4, 6, 5]
#printed lists while running the program
But the same problem still occurs everytime when I check the timelst after executing the program.
I used python 3.7 with spyder. There should not be a problem with my compiler, as I have tried running this on ipython and I still got the same result.
Can anyone please help me solving this problem? Thanks!
python list
python list
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
edited Nov 19 at 23:11
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
asked Nov 19 at 21:42
PotatoPoweredLaptop
64
64
1
Multiple nested loops like this is generally considered bad programming practice, especially when the loops do the same thing. It's not entirely clear to me what you want the output to be ... what is the intention of this code?
– Jonah Bishop
Nov 19 at 22:04
Basically I want to append every possible combinations of the numbers under six to a list, such as 123456 123465 123546 123564 ... 654312 654321, and this is the most efficient way I can think to achieve this, rather than append all possible lists and remove the ones with repeating numbers. The program is not totally finished, it should work, ideally and maybe logically, if I add timelst.pop() at the end of every loop after the previous loop finishes. I did try it, but it gave me a much worse result like [[1], [1], [1], ...., [6], [6], [6]]. I realise maybe they are caused by the same problem.
– PotatoPoweredLaptop
Nov 19 at 22:30
check the indentation of yourcheckfunction.
– Paul H
Nov 19 at 22:56
Sorry that was just a silly mistake when I was transferring my code to my post. It was correct in my compiler and that did not cause any problem, and the indentation of function has been corrected in my post
– PotatoPoweredLaptop
Nov 19 at 23:20
add a comment |
1
Multiple nested loops like this is generally considered bad programming practice, especially when the loops do the same thing. It's not entirely clear to me what you want the output to be ... what is the intention of this code?
– Jonah Bishop
Nov 19 at 22:04
Basically I want to append every possible combinations of the numbers under six to a list, such as 123456 123465 123546 123564 ... 654312 654321, and this is the most efficient way I can think to achieve this, rather than append all possible lists and remove the ones with repeating numbers. The program is not totally finished, it should work, ideally and maybe logically, if I add timelst.pop() at the end of every loop after the previous loop finishes. I did try it, but it gave me a much worse result like [[1], [1], [1], ...., [6], [6], [6]]. I realise maybe they are caused by the same problem.
– PotatoPoweredLaptop
Nov 19 at 22:30
check the indentation of yourcheckfunction.
– Paul H
Nov 19 at 22:56
Sorry that was just a silly mistake when I was transferring my code to my post. It was correct in my compiler and that did not cause any problem, and the indentation of function has been corrected in my post
– PotatoPoweredLaptop
Nov 19 at 23:20
1
1
Multiple nested loops like this is generally considered bad programming practice, especially when the loops do the same thing. It's not entirely clear to me what you want the output to be ... what is the intention of this code?
– Jonah Bishop
Nov 19 at 22:04
Multiple nested loops like this is generally considered bad programming practice, especially when the loops do the same thing. It's not entirely clear to me what you want the output to be ... what is the intention of this code?
– Jonah Bishop
Nov 19 at 22:04
Basically I want to append every possible combinations of the numbers under six to a list, such as 123456 123465 123546 123564 ... 654312 654321, and this is the most efficient way I can think to achieve this, rather than append all possible lists and remove the ones with repeating numbers. The program is not totally finished, it should work, ideally and maybe logically, if I add timelst.pop() at the end of every loop after the previous loop finishes. I did try it, but it gave me a much worse result like [[1], [1], [1], ...., [6], [6], [6]]. I realise maybe they are caused by the same problem.
– PotatoPoweredLaptop
Nov 19 at 22:30
Basically I want to append every possible combinations of the numbers under six to a list, such as 123456 123465 123546 123564 ... 654312 654321, and this is the most efficient way I can think to achieve this, rather than append all possible lists and remove the ones with repeating numbers. The program is not totally finished, it should work, ideally and maybe logically, if I add timelst.pop() at the end of every loop after the previous loop finishes. I did try it, but it gave me a much worse result like [[1], [1], [1], ...., [6], [6], [6]]. I realise maybe they are caused by the same problem.
– PotatoPoweredLaptop
Nov 19 at 22:30
check the indentation of your
check function.– Paul H
Nov 19 at 22:56
check the indentation of your
check function.– Paul H
Nov 19 at 22:56
Sorry that was just a silly mistake when I was transferring my code to my post. It was correct in my compiler and that did not cause any problem, and the indentation of function has been corrected in my post
– PotatoPoweredLaptop
Nov 19 at 23:20
Sorry that was just a silly mistake when I was transferring my code to my post. It was correct in my compiler and that did not cause any problem, and the indentation of function has been corrected in my post
– PotatoPoweredLaptop
Nov 19 at 23:20
add a comment |
2 Answers
2
active
oldest
votes
You can make your life a lot easier by using the itertools.permutations() function call:
import itertools
mylist = [1, 2, 3, 4, 5, 6]
for item in itertools.permutations(mylist):
# Do something with item, which is a permutation of 'mylist'
print(item)
Note, however, that this results in a big (6! ... that is six-factorial) list of results. If you only want combinations, take a look at itertools.combinations() instead.
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
add a comment |
The problem has been solved! I should have realised earlier that there had been an issue about the shallow and deep copy of the list I have appended. I have add
usednum = copy.deepcopy(usednum)
to the final loop, so it now looks like
for i6 in range(1, 7):
if not check(i6, usednum):
continue
else:
timelst.append(usednum)
usednum = copy.deepcopy(usednum) #the new copy of the appended list, so now there's nothing to do with the appended one.
print('i6',usednum.pop())
break
usednum.pop()
which can work properly now
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can make your life a lot easier by using the itertools.permutations() function call:
import itertools
mylist = [1, 2, 3, 4, 5, 6]
for item in itertools.permutations(mylist):
# Do something with item, which is a permutation of 'mylist'
print(item)
Note, however, that this results in a big (6! ... that is six-factorial) list of results. If you only want combinations, take a look at itertools.combinations() instead.
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
add a comment |
You can make your life a lot easier by using the itertools.permutations() function call:
import itertools
mylist = [1, 2, 3, 4, 5, 6]
for item in itertools.permutations(mylist):
# Do something with item, which is a permutation of 'mylist'
print(item)
Note, however, that this results in a big (6! ... that is six-factorial) list of results. If you only want combinations, take a look at itertools.combinations() instead.
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
add a comment |
You can make your life a lot easier by using the itertools.permutations() function call:
import itertools
mylist = [1, 2, 3, 4, 5, 6]
for item in itertools.permutations(mylist):
# Do something with item, which is a permutation of 'mylist'
print(item)
Note, however, that this results in a big (6! ... that is six-factorial) list of results. If you only want combinations, take a look at itertools.combinations() instead.
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
You can make your life a lot easier by using the itertools.permutations() function call:
import itertools
mylist = [1, 2, 3, 4, 5, 6]
for item in itertools.permutations(mylist):
# Do something with item, which is a permutation of 'mylist'
print(item)
Note, however, that this results in a big (6! ... that is six-factorial) list of results. If you only want combinations, take a look at itertools.combinations() instead.
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
answered Nov 19 at 22:55
Jonah Bishop
8,26732957
8,26732957
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
add a comment |
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
Thank you very much! I didn't realise I could find a toolkit to reach my purpose. However, at this stage, I am still wondering what makes my program not work properly, as the 'usednum' list printed in the previous step is [1, 2, 3, 4, 6, 5], but as it is appended to the 'timelst', it becomes [1, 2, 3, 4, 5, 6].
– PotatoPoweredLaptop
Nov 19 at 23:17
add a comment |
The problem has been solved! I should have realised earlier that there had been an issue about the shallow and deep copy of the list I have appended. I have add
usednum = copy.deepcopy(usednum)
to the final loop, so it now looks like
for i6 in range(1, 7):
if not check(i6, usednum):
continue
else:
timelst.append(usednum)
usednum = copy.deepcopy(usednum) #the new copy of the appended list, so now there's nothing to do with the appended one.
print('i6',usednum.pop())
break
usednum.pop()
which can work properly now
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
add a comment |
The problem has been solved! I should have realised earlier that there had been an issue about the shallow and deep copy of the list I have appended. I have add
usednum = copy.deepcopy(usednum)
to the final loop, so it now looks like
for i6 in range(1, 7):
if not check(i6, usednum):
continue
else:
timelst.append(usednum)
usednum = copy.deepcopy(usednum) #the new copy of the appended list, so now there's nothing to do with the appended one.
print('i6',usednum.pop())
break
usednum.pop()
which can work properly now
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
add a comment |
The problem has been solved! I should have realised earlier that there had been an issue about the shallow and deep copy of the list I have appended. I have add
usednum = copy.deepcopy(usednum)
to the final loop, so it now looks like
for i6 in range(1, 7):
if not check(i6, usednum):
continue
else:
timelst.append(usednum)
usednum = copy.deepcopy(usednum) #the new copy of the appended list, so now there's nothing to do with the appended one.
print('i6',usednum.pop())
break
usednum.pop()
which can work properly now
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
The problem has been solved! I should have realised earlier that there had been an issue about the shallow and deep copy of the list I have appended. I have add
usednum = copy.deepcopy(usednum)
to the final loop, so it now looks like
for i6 in range(1, 7):
if not check(i6, usednum):
continue
else:
timelst.append(usednum)
usednum = copy.deepcopy(usednum) #the new copy of the appended list, so now there's nothing to do with the appended one.
print('i6',usednum.pop())
break
usednum.pop()
which can work properly now
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
answered Nov 19 at 23:58
PotatoPoweredLaptop
64
64
add a comment |
add a comment |
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1
Multiple nested loops like this is generally considered bad programming practice, especially when the loops do the same thing. It's not entirely clear to me what you want the output to be ... what is the intention of this code?
– Jonah Bishop
Nov 19 at 22:04
Basically I want to append every possible combinations of the numbers under six to a list, such as 123456 123465 123546 123564 ... 654312 654321, and this is the most efficient way I can think to achieve this, rather than append all possible lists and remove the ones with repeating numbers. The program is not totally finished, it should work, ideally and maybe logically, if I add timelst.pop() at the end of every loop after the previous loop finishes. I did try it, but it gave me a much worse result like [[1], [1], [1], ...., [6], [6], [6]]. I realise maybe they are caused by the same problem.
– PotatoPoweredLaptop
Nov 19 at 22:30
check the indentation of your
checkfunction.– Paul H
Nov 19 at 22:56
Sorry that was just a silly mistake when I was transferring my code to my post. It was correct in my compiler and that did not cause any problem, and the indentation of function has been corrected in my post
– PotatoPoweredLaptop
Nov 19 at 23:20