I have a list, num_list, which contains random numbers identified by using random.
One for loop is responsible for selecting numbers from the original list and my compare function is supposed to compare the obtained number with the numbers present in my new list. If the number is greater than the first number obtained from the for loop, I check if the same num is less than the next num in the new list. This means that the num is between the numbers and is therefore inserted there. However, I am stuck in a loop and can't understand what is going on. I have tried everything I know but, sadly, I'm still just a beginner. I hope someone here can make head or tail of this unfortunate mess.
new_list.append((num_list)[0])
def compare(num, new_list, index, temp_char):
for char in new_list:
print(char)
if num > char:
index = new_list.index(char)
try:
temp_char = (new_list)[index + 1]
if num <= temp_char:
new_list.insert(index+1, num)
except:
new_list.insert(index, num)
break
for num in num_list:
#print(num)
compare(num, new_list, index, temp_char)
print(new_list)
As Sayse said in comment, you shouldn't modify a list while iterating over it. This leads to undefined behavior. Instead, copy the list new_list before iterating over it. To copy the list, there are many solutions, but here's a simple one: copied = [i for i in new_list]
This won't work if your list is a nested list.
Related
There is a code:
list = [1, 2]
while list[-1]+list[-2] <= 4000000:
list.append(list[-1] + list[-2])
for i in list:
if i % 2 == 1:
print(i)
list.remove(i)
print(list)
print(sum(list))
You shouldn't modify a list (or any container) while iterating through it.
One way to go around it is to use another container,
in_list = [1, 2]
while in_list[-1]+in_list[-2] <= 20:
in_list.append(in_list[-1] + in_list[-2])
print(in_list)
out_list = []
for i in in_list:
if i % 2 != 1:
print(i)
out_list.append(i)
print(out_list)
print(sum(out_list))
This code uses a different approach than yours: it creates the input list, then while iterating it adds the even elements to a new, output list. This has the same effect as removing the odd elements from the input list, however, it doesn't break the iteration by modifying the input list.
Like said in the comments, you shouldn't use built-in names ("list") for your variable names - it will shadow them. Also, when you develop and debug your code it's best to stick to smaller examples. Here I use 20 instead of 4,000,000 - much easier to track and doesn't lose the meaning.
Found that the control never reaches the inner while loop in this snippet for deleting a duplicate number
numbers=[1,6,6,7]
k=len(numbers)
i=0
j=0
while i in range(k-1):
while j in range(i+1,k):
if numbers[i] == numbers[j]:
numbers.remove(numbers[j])
k-=1
j-=1
j += 1
i += 1
print(numbers)
Your code does not make j start at i+1. Instead it starts at zero and never changes. The inner loop never runs because 0 is outside of the range you are testing.
Try this simple change:
i=0
while i < k+1:
j=i+1
while j < k:
if numbers[i] == numbers[j]:
...
The main change is moving the initialization of j inside the first while loop, so it updates each time you go through it, and never starts out less than or equal to i.
The other change I made is much less important. Rather than using i in range(...) and j in range(...) for the while loop conditions, I just did an inequality test. This is exactly the same as what the range membership test does under the covers, but avoids unnecessary testing for things that can't happen (like j being too small, now). It also makes the loop look a lot less like a for loop, which uses for i in range(...) syntax a lot (with a different meaning).
Another issue you may run into later, with some sets with multiple sets of duplicates is that your code to remove the jth element probably doesn't do what you intend. The call numbers.remove(numbers[j]) removes the first value equal to numbers[j] from the list, which is going to be the one at index i rather than the one at index j. To delete a list item by index, you want to use del numbers[j].
It doesn't reach because j and i starts at 0 value and in the inner while loop the condition is j in range(i+1, k) which means range(1, 4) and 0 in range(1, 4) would be False. Anyways, you should avoid using j and i as counters and use a for loop instead.
But the solution is easier and doesn't need to traverse the list, if you wanna remove the duplicate values, you can do as below:
numbers = [1, 6, 6, 7]
print(list(set(numbers)))
The result is: [1, 6, 7]
You could remove duplicates from a list in Python by using the dict.fromkeys().
numbers=[1,6,6,7]
final_numbers = list(dict.fromkeys(numbers))
print(final_numbers)
In this example, we use the dict.fromkeys() method to create a dictionary from numbers variable. We then use list() to convert our data from a dictionary back to a list. Then, on the final line, we print out our revised list.
Another option is to use set.
Sets are used to store collections of unique items in Python. Unlike lists, sets cannot store duplicate values.
We can convert our list to a set to remove duplicated items.
numbers=[1,6,6,7]
final_numbers = list(set(numbers))
print(final_numbers)
This question already has answers here:
Are Python variables pointers? Or else, what are they?
(9 answers)
Closed 3 years ago.
I am writing a little program to create a list of permutations. I read about the algorithm on wikipedia.
My algorithm basically takes an initially sorted list of numbers, and permutes it in place. It then appends this new permutation to a list. When all permutations are found, it returns the list of lists containing all the permutations. It is very good at printing out the expected results, but when I try to add those results to a list, things get a little funny.
I noticed that every time I find the next permutation and append it to, the previous list elements get updated to the new permutation. So, at the end of it all, what gets returned is a list containing a bunch of copies of the same permutation (exactly the last permutation).
I've read that Python is pass by value, pass by reference and I've also read that it's neither. I'm not smart enough to argue with any of those people, but I am wondering why my program is doing this, and how to remedy it:
def lexi_order(nums):
permutations = []
length = len(nums)
while True:
# find largest index i such that nums[i] < nums[i + 1]
exists = False
for j, elem in enumerate(nums):
# check if last element
if j == length - 1:
break
if elem < nums[j + 1]:
i = j
exists = True
if not exists:
break
# find largest index k, such that i < k AND nums[i] < nums[k]
for j in range(i + 1, length):
if nums[j] > nums[i]:
k = j
# swap order of nums[i] and nums[k]
nums[i], nums[k] = nums[k], nums[i]
# reverse order of elements starting at position i+1
to_reverse = nums[i+1:][::-1]
nums[i+1::] = to_reverse
permutations.append(nums)
print(permutations)
return permutations
You're modifying the input (nums) in place each iteration through the loop, and then you keep adding a reference to the input to permutations. To fix it, make a copy of nums at the beginning of the loop and use it instead of the original everywhere inside it.
When you append nums to permutations, you are appending a reference to it, not copying all of the data over. When you modify nums, it gets modified everywhere. Python is pass by reference. If you make a change to a variable (not to be confused with reassigning it), that change will be reflected everywhere.
You need to make a copy of the passed nums, otherwise you are working on the passed reference. E.g.
def lexi_order(nums):
permutations = []
nums = list(nums) # We are now working on a copy, and won't mutate the original whatsoever.
length = len(nums)
...
So I wrote this algorithm where given a set of integers it will remove all integers except 0 and 7 and then it will check if the remaining integers are in a certain order and then will return a boolean. Code below:
def spy_game(nums):
for i in nums:
if i != 0:
if i == 7:
continue
else:
nums.remove(i)
else:
continue
stringlist = [str(o) for o in nums]
mystring = ''.join(stringlist)
return '007' in mystring
spy_game([1,0,2,4,0,7,5])
Now the problem is that if I run
(for example) spy_game([1,0,2,4,0,7,5]) it will not return True regardless of the fact that the sequence of interest is present. After I decided to return the list per se after the filtration process, I found that all numbers except the ones in the middle got filtered out. So in this example, if I return nums it will return [0, 4, 0, 7] although the 4 should've been removed. I am aware that there are more optimal alternatives to this algorithm but I just want to understand why it doesn't work. Thank you.
Instead of modifying the list, use another list to keep track of the wanted numbers.
You should not modify the list while iterating on it.
Here's a cleaned up version
def spy_game(nums):
ans = []
for i in nums:
if i == 0 or i == 7:
ans.append(i)
stringlist = [str(o) for o in ans]
mystring = ''.join(stringlist)
return '007' in mystring
zenwraight's comment says what the problem is: in Python, you can't modify a list while iterating over it.
As for why, the Python documentation discusses this in a note on the for statement's section:
An internal counter is used to keep track of which item is used next, and this is incremented on each iteration. … This means that if the [loop body] deletes the current … item from the sequence, the next item will be skipped (since it gets the index of the current item which has already been treated).
The documentation also describes what happens when you insert an element during a loop, and suggests one possible solution (using a slice to copy the list: for i in nums[:]: ...). In your use case, that solution is likely to work fine, but it is considerably less efficient than options that don't copy the entire list.
A better solution might be to use another list comprehension:
nums = [i for i in nums if i == 0 or i == 7]
I have written a simple python program
l=[1,2,3,0,0,1]
for i in range(0,len(l)):
if l[i]==0:
l.pop(i)
This gives me error 'list index out of range' on line if l[i]==0:
After debugging I could figure out that i is getting incremented and list is getting reduced.
However, I have loop termination condition i < len(l). Then why I am getting such error?
You are reducing the length of your list l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.
It looks like what you want to do is:
l = [x for x in l if x != 0]
which will return a copy of l without any of the elements that were zero (that operation is called a list comprehension, by the way). You could even shorten that last part to just if x, since non-zero numbers evaluate to True.
There is no such thing as a loop termination condition of i < len(l), in the way you've written the code, because len(l) is precalculated before the loop, not re-evaluated on each iteration. You could write it in such a way, however:
i = 0
while i < len(l):
if l[i] == 0:
l.pop(i)
else:
i += 1
The expression len(l) is evaluated only one time, at the moment the range() builtin is evaluated. The range object constructed at that time does not change; it can't possibly know anything about the object l.
P.S. l is a lousy name for a value! It looks like the numeral 1, or the capital letter I.
You're changing the size of the list while iterating over it, which is probably not what you want and is the cause of your error.
Edit: As others have answered and commented, list comprehensions are better as a first choice and especially so in response to this question. I offered this as an alternative for that reason, and while not the best answer, it still solves the problem.
So on that note, you could also use filter, which allows you to call a function to evaluate the items in the list you don't want.
Example:
>>> l = [1,2,3,0,0,1]
>>> filter(lambda x: x > 0, l)
[1, 2, 3]
Live and learn. Simple is better, except when you need things to be complex.
What Mark Rushakoff said is true, but if you iterate in the opposite direction, it is possible to remove elements from the list in the for-loop as well. E.g.,
x = [1,2,3,0,0,1]
for i in range(len(x)-1, -1, -1):
if x[i] == 0:
x.pop(i)
It's like a tall building that falls from top to bottom: even if it is in the middle of collapse, you still can "enter" into it and visit yet-to-be-collapsed floors.
I think the best way to solve this problem is:
l = [1, 2, 3, 0, 0, 1]
while 0 in l:
l.remove(0)
Instead of iterating over list I remove 0 until there aren't any 0 in list
List comprehension will lead you to a solution.
But the right way to copy a object in python is using python module copy - Shallow and deep copy operations.
l=[1,2,3,0,0,1]
for i in range(0,len(l)):
if l[i]==0:
l.pop(i)
If instead of this,
import copy
l=[1,2,3,0,0,1]
duplicate_l = copy.copy(l)
for i in range(0,len(l)):
if l[i]==0:
m.remove(i)
l = m
Then, your own code would have worked.
But for optimization, list comprehension is a good solution.
The problem was that you attempted to modify the list you were referencing within the loop that used the list len(). When you remove the item from the list, then the new len() is calculated on the next loop.
For example, after the first run, when you removed (i) using l.pop(i), that happened successfully but on the next loop the length of the list has changed so all index numbers have been shifted. To a certain point the loop attempts to run over a shorted list throwing the error.
Doing this outside the loop works, however it would be better to build and new list by first declaring and empty list before the loop, and later within the loop append everything you want to keep to the new list.
For those of you who may have come to the same problem.
I am using python 3.3.5. The above solution of using while loop did not work for me. Even if i put print (i) after len(l) it gave me an error. I ran the same code in command line (shell)[ window that pops up when we run a function] it runs without error. What i did was calculated len(l) outside the function in main program and passed the length as a parameter. It worked. Python is weird sometimes.
I think most solutions talk here about List Comprehension, but if you'd like to perform in place deletion and keep the space complexity to O(1); The solution is:
i = 0
for j in range(len(arr)):
if (arr[j] != 0):
arr[i] = arr[j]
i +=1
arr = arr[:i]
x=[]
x = [int(i) for i in input().split()]
i = 0
while i < len(x):
print(x[i])
if(x[i]%5)==0:
del x[i]
else:
i += 1
print(*x)
Code:
while True:
n += 1
try:
DATA[n]['message']['text']
except:
key = DATA[n-1]['message']['text']
break
Console :
Traceback (most recent call last):
File "botnet.py", line 82, in <module>
key =DATA[n-1]['message']['text']
IndexError: list index out of range
I recently had a similar problem and I found that I need to decrease the list index by one.
So instead of:
if l[i]==0:
You can try:
if l[i-1]==0:
Because the list indices start at 0 and your range will go just one above that.