I have the following code.
for idx in range(len(networks)):
net_ = networks[idx]
lastId=0
for layerUptID in range(len(net_[1])):
retNet,lastId=cn_.UpdateTwoConvLayers(deepcopy(net_),lastId)
networks.append(retNet)
if(lastId==-1):
break
networks has only one net at the beginning.
After running the line retNet,lastId=cn_.UpdateTwoConvLayers(deepcopy(net_),lastId), I have additional six nets and appended to networks.
So after this lastId ==-1, go back to first for loop with len(networks) is 7.
For the next idx, idx=1 and continue.
Then, len(networks) is 13. Then go back to first for loop.
After this, the first for loop breaks.
I am expecting to continue for idx is 2, but it breaks.
What could be the issue?
If you try using a WHILE loop instead of FOR loop, the break statement would be check if the loop is on the last item in 'networks' collection.
This way the network length would be calculated in each loop iteration
For starters: Iterating, or looping, over the list (or data) you're editing is bad practice. Keep that in mind while coding.
This means if you plan to edit what you're looping on, in your case networks, then you're going to have a bad time looping over it. I would advise to break it up into two code parts:
The first part creates a new list of whatever it is you want WHILE looping.
The second part replaces the list you've used to generate what you wanted.
Another thing which could go wrong is net_[i] may not be set up for some i, and you're trying to access it here:
for layerUptID in range(len(net_[1])):
What if there is nothing in net_[1]?
To avoid these errors, usually verifying your data is a great way to start. If it is not null, then proceed, otherwise, print an error.
This is what I can think of. Hope it helps.
If I understood correctly your problem is that you've added new elements to networks, i.e. have increased length of networks and expect that for-loop will pick up this changes, well it's not, let's look at following snippet
elements = [1]
indices = range(len(elements))
for index in indices:
print('index is', index)
elements.append(2)
print('elements count is', len(elements))
print('indices count is', len(indices))
outputs are
index is 0
elements count is 2
indices count is 1
so as we can see despite the fact that length of elements list has changed, range object which is used in for-loop has not. This happens because len returns int object which are immutable, so when you change list length its length becomes different object and range function has no idea about this changes.
Finally, we can use while loop here like
while networks:
net_ = networks.pop()
lastId = 0
for layerUptID in range(len(net_[1])):
retNet, lastId = cn_.UpdateTwoConvLayers(deepcopy(net_), lastId)
networks.append(retNet)
if lastId == -1:
break
Related
I'm trying to enter a 100 number square into a 2D list https://i.stack.imgur.com/eSFiO.png with each list containing 10 numbers 1-10,11-20,21-30 and so on. This is my code so far but when I run it in my editor it just keeps running and eventually crashes without printing anything. Please explain to me what I am doing wrong. Thanks
number_square=[[],[],[],[],[],[],[],[],[],[]]
number_list=[1,2,3,4,5,6,7,8,9,10]
for row in number_square:
for number in number_list:
number_square.append(number)
number_list.remove(number)
number_list.append(number+10)
print(number_square)
That's because you aren't accessing the content of neither row nor number in the for loops. Here is a suggestion:
number_square=[[],[],[],[],[],[],[],[],[],[]]
number_list=[1,2,3,4,5,6,7,8,9,10]
i = 0
for row in number_square:
for i in range(len(number_list)):
row.append(number_list[i])
number_list[i] += 10
print(number_square)
Note that the first loop is equivalent to for each. In this syntax, you can't alter the value of the item in a list. In the second loop, I put a "traditional" for loop with range to alter the values in number_list.
There are many changes required in your code. Good effort from your side on trying the code. I have modified the code and probably you can get the logic from it.
number_square=[[],[],[],[],[],[],[],[],[],[]]
number_list=[10,20,30,40,50,60,70,80,90,100]
for i in range(0,len(number_square)):
for j in range(number_list[i]-9,number_list[i]+1):
number_square[i].append(j)
Problems:
Removing from number_list while iterating over it
Appending to number_square instead of row
If I were you, I'd use a list comprehension with ranges instead:
number_square = [list(range(i, i+10)) for i in range(1, 101, 10)]
Due credit to Onyambu for suggesting something similar in a comment
This question already has answers here:
How to test multiple variables for equality against a single value?
(31 answers)
Closed 6 years ago.
import os
os.chdir('G:\\f5_automation')
r = open('G:\\f5_automation\\uat.list.cmd.txt')
#print(r.read().replace('\n', ''))
t = r.read().split('\n')
for i in range(len(t)):
if ('inherited' or 'device-group' or 'partition' or 'template' or 'traffic-group') in t[i]:
t.pop(i)
print(i,t[i])
In the above code, I get an index error at line 9: 'if ('inherited' or 'device-group'...etc.
I really don't understand why. How can my index be out of range if it's the perfect length by using len(t) as my range?
The goal is to pop any indexes from my list that contain any of those substrings. Thank you for any assistance!
This happens because you are editing the list while looping through it,
you first get the length which is 10 for example, then you loop through the thing 10 times. but as soon as you've deleted one thing the list will only be 9 long.
A way around this is to create a new list of things you want to keep and use that one instead.
I've slightly edited your code and done something similar.
t = ['inherited', 'cookies', 'device-group']
interesing_things = []
for i in t:
if i not in ['inherited', 'device-group', 'partition', 'template', 'traffic-group']:
interesing_things.append(i)
print(i)
Let's say len(t) == 5.
We'll process i taking values [0,1,2,3,4]
After we process i = 0, we pop one value from t. len(t) == 4 now. This would mean error if we get to i = 4. However, we're still going to try to go up to 4 because our range is already inited to be up to 4.
Next (i = 1) step ensures an error on i = 3.
Next (i = 2) step ensures an error on i = 2, but that is already processed.
Next (i = 3) step yields an error.
Instead, you should do something like this:
while t:
element = t.pop()
print(element)
On a side note, you should replace that in check with sets:
qualities_we_need = {'inherited', 'device-group', 'partition'} # put all your qualities here
And then in loop:
if qualities_we_need & set(element):
print(element)
If you need indexes you could either use one more variable to keep track of index of value we're currently processing, or use enumerate()
As many people said in the comments, there are several problems with your code.
The or operator sees the values on its left and right as booleans and returns the first one that is True (from left to right). So your parenthesis evaluates to 'inherited' since any non-empty string is True. As a result, even if your for loop was working, you would be popping elements that are equal to 'inherited' only.
The for loop is not working though. That happens because the size of the list you are iterating over is changing as you loop through and you will get an index-out-of-range error if an element of the list is actually equal to 'inherited' and gets popped.
So, take a look at this:
import os
os.chdir('G:\\f5_automation')
r = open('G:\\f5_automation\\uat.list.cmd.txt')
print(r.read().replace('\n', ''))
t = r.read().split('\n')
t_dupl = t[:]
for i, items in enumerate(t_dupl):
if items in ['inherited', 'device-group', 'partition', 'template', 'traffic-group']:
print(i, items)
t.remove(items)
By duplicating the original list, we can use its items as a "pool" of items to pick from and modify the list we are actually interested in.
Finally, know that the pop() method returns the item it removes from the list and this is something you do not need in your example. remove() works just fine for you.
As a side note, you can probably replace your first 5 lines of code with this:
with open('G:\\f5_automation\\uat.list.cmd.txt', 'r') as r:
t = r.readlines()
the advantage of using the with statement is that it automatically handles the closing of the file by itself when the reading is done. Finally, instead of reading the whole file and splitting it on linebreaks, you can just use the built-in readlines() method which does exactly that.
I'm taking a course on programming (I'm a complete beginner) and the current assignment is to create a Python script that sorts a list of numbers in ascending order without using built-in functions like "sorted".
The script I started to come up with is probably laughably convoluted and inefficient, but I'd like to try to make it work eventually. In other words, I don't want to just copy someone else's script that's better.
Also, as far as I can tell, this script (if it ever functioned) would probably just put things in a NEW order that wouldn't necessarily be ascending. This is just a start, though, and hopefully I'll fix that later.
Anyway, I've run in to several problems with the current incarnation, and the latest is that it just runs forever without printing anything.
So here it is (with hashes explaining what I was trying to accomplish). If someone could look over it and tell me why the code does not match my explanations of what each block is supposed to do, that would be great!
# The numbers to be inputted, could be anything
numList = [1, 25, 5, 6, 17, 4]
# The final (hopefully sorted) list
numSort = []
# The index
i = 0
# Run the loop until you run out of numbers
while len(numList) != 0:
# If there's only one value left in numList,
# attach it to the end of numSort.
# (Variable 'basket' is just for transporting numbers)
if len(numList) == 1:
basket = numList.pop()
numSort.append(basket)
# The rest of the elifs are supposed to compare values
# that sit next to each other.
# If there's still a number in numList after 'i'
# and 'i' is smaller than that next number
# then pop 'i' and attach it to the end of numSort
elif numList[i+1] != numList[-1] and numList[i] < numList[i+1]:
basket = numList.pop(i)
numSort.append(basket)
# If there's NOT a number after 'i'
# then compare 'i' to the first number in the list instead.
elif numList[i+1] == numList[-1] and numList[i] < numList[0]:
basket = numList.pop(i)
numSort.append(basket)
# If 'i' IS the last number in the list
# and has nothing to compare itself to,
# Then start over and go through it again
# from the beginning
elif numList [i+1] == numList[-1]:
i = 0
# If 'i' is not at the end of numList yet
# and 'i' is NOT smaller than the next number
# and there are still numbers left
# then move on to the next pair
# and continue comparing and moving numbers
else:
i = i+1
# Hopefully these will be in ascending order eventually.
print(numSort)
Here is a simple way to sort your list with a classic loop :
myList = [2,99,0,56,8,1]
for i,value in enumerate(myList):
for j,innerValue in enumerate(myList):
if value < innerValue: #for order desc use '>'
myList[j],myList[i]=myList[i],myList[j]
print(myList)
The Algorithm behind this code is :
fix one value of the list and compare it with the rest
if it is smallest then switch the index of two values
I hope this will help you
Your conditions are essentially:
If there is only one number in the list
The current number is less than the next, which is not equal to the last number
The current number is less than the first number, and the next number is equal to the last number
The next number is equal to the last number
If you trace out the code by hand you will see how in many cases, none of these will evaluate to true and your "else" will be executed and i will be incremented. Under these conditions, certain numbers will never be removed from your original list (none of your elifs will catch them), i will increment until the next number is equal to the last number, i will be reset to zero, repeat. You are stuck in an infinite loop. You will need to update your if-statement in such a way that all numbers will eventually be caught by a block other than your final elif.
On a separate note, you are potentially comparing a number to only one number in the current list before appending it to the "sorted" list. You will likely need to compare the number you want to add to your "sorted" list and find its proper place in THAT list rather than merely appending.
You should also consider finding the end of list using a method more like
if i == len(numList) - 1
This will compare the index to the length of the list rather than comparing more values in the list which are not necessarily relevant to the order you are trying to create.
I am writing a code for a class that wants me to make a code to check the substring in a string using nested loops.
Basically my teacher wants to prove how the function 'in', as in:
ana in banana will return True.
The goal of the program is to make a function of 2 parameters,
substring(subStr,fullStr)
that will print out a sentence saying if subStr is a substring of fullStr, my program is as follows:
def substring(subStr,fullStr):
tracker=""
for i in (0,(len(fullStr)-1)):
for j in (0,(len(subStr)-1)):
if fullStr[i]==subStr[j]:
tracker=tracker+subStr[j]
i+=1
if i==(len(fullStr)-1):
break
if tracker==subStr:
print "Yes",subStr,"is a substring of",fullStr
When i called the function in the interpreter 'substring("ana","banana")', it printed out a traceback error on line 5 saying string index out of range:
if fullStr[i]==subStr[j]:
I'm banging my head trying to find the error. Any help would be appreciated
There are a few separate issues.
You are not reseting tracker in every iteration of the outer loop. This means that the leftovers from previous iterations contaminate later iterations.
You are not using range, and are instead looping over a tuple of just the 0 and the length of each string.
You are trying to increment the outer counter and skipping checks for the iteration of the outer loop.
You are not doing the bounds check correctly before trying to index into the outer string.
Here is a corrected version.
def substring(subStr,fullStr):
for i in range(0,(len(fullStr))):
tracker=""
for j in range(0,(len(subStr))):
if i + j >= len(fullStr):
break
if fullStr[i+j]==subStr[j]:
tracker=tracker+subStr[j]
if tracker==subStr:
print "Yes",subStr,"is a substring of",fullStr
return
substring("ana", "banana")
First off, your loops should be
for i in xrange(0,(len(fullStr))):
for example. i in (0, len(fullStr)-1) will have i take on the value of 0 the first time around, then take on len(fullStr)-1 the second time. I assume by your algorithm you want it to take on the intermediate values as well.
Now as for the error, consider i on the very last pass of the for loop. i is going to be equal to len(fullStr)-1. Now when we execute i+=1, i is now equal to len(fullStr). This does not fufill the condition of i==len(fullStr)-1, so we do not break, we loop, and we crash. It would be better if you either made it if i>=len(fullStr)-1 or checked for i==len(fullStr)-1 before your if fullStr[i]==subStr[j]: statement.
Lastly, though not related to the question specifically, you do not reset tracker each time you stop checking a certain match. You should place tracker = "" after the for i in xrange(0,(len(fullStr))): line. You also do not check if tracker is correct after looping through the list starting at i, nor do you break from the loop when you get a mismatch(instead continuing and possibly picking up more letters that match, but not consecutively.)
Here is a fully corrected version:
def substring(subStr,fullStr):
for i in xrange(0,(len(fullStr))):
tracker="" #this is going to contain the consecutive matches we find
for j in xrange(0,(len(subStr))):
if i==(len(fullStr)): #end of i; no match.
break
if fullStr[i]==subStr[j]: #okay, looks promising, check the next letter to see if it is a match,
tracker=tracker+subStr[j]
i+=1
else: #found a mismatch, leave inner loop and check what we have so far.
break
if tracker==subStr:
print "Yes",subStr,"is a substring of",fullStr
return #we already know it is a substring, so we don't need to check the rest
I know that it is not allowed to remove elements while iterating a list, but is it allowed to add elements to a python list while iterating. Here is an example:
for a in myarr:
if somecond(a):
myarr.append(newObj())
I have tried this in my code and it seems to work fine, however I don't know if it's because I am just lucky and that it will break at some point in the future?
EDIT: I prefer not to copy the list since "myarr" is huge, and therefore it would be too slow. Also I need to check the appended objects with "somecond()".
EDIT: At some point "somecond(a)" will be false, so there can not be an infinite loop.
EDIT: Someone asked about the "somecond()" function. Each object in myarr has a size, and each time "somecond(a)" is true and a new object is appended to the list, the new object will have a size smaller than a. "somecond()" has an epsilon for how small objects can be and if they are too small it will return "false"
Why don't you just do it the idiomatic C way? This ought to be bullet-proof, but it won't be fast. I'm pretty sure indexing into a list in Python walks the linked list, so this is a "Shlemiel the Painter" algorithm. But I tend not to worry about optimization until it becomes clear that a particular section of code is really a problem. First make it work; then worry about making it fast, if necessary.
If you want to iterate over all the elements:
i = 0
while i < len(some_list):
more_elements = do_something_with(some_list[i])
some_list.extend(more_elements)
i += 1
If you only want to iterate over the elements that were originally in the list:
i = 0
original_len = len(some_list)
while i < original_len:
more_elements = do_something_with(some_list[i])
some_list.extend(more_elements)
i += 1
well, according to http://docs.python.org/tutorial/controlflow.html
It is not safe to modify the sequence
being iterated over in the loop (this
can only happen for mutable sequence
types, such as lists). If you need to
modify the list you are iterating over
(for example, to duplicate selected
items) you must iterate over a copy.
You could use the islice from itertools to create an iterator over a smaller portion of the list. Then you can append entries to the list without impacting the items you're iterating over:
islice(myarr, 0, len(myarr)-1)
Even better, you don't even have to iterate over all the elements. You can increment a step size.
In short: If you'are absolutely sure all new objects fail somecond() check, then your code works fine, it just wastes some time iterating the newly added objects.
Before giving a proper answer, you have to understand why it considers a bad idea to change list/dict while iterating. When using for statement, Python tries to be clever, and returns a dynamically calculated item each time. Take list as example, python remembers a index, and each time it returns l[index] to you. If you are changing l, the result l[index] can be messy.
NOTE: Here is a stackoverflow question to demonstrate this.
The worst case for adding element while iterating is infinite loop, try(or not if you can read a bug) the following in a python REPL:
import random
l = [0]
for item in l:
l.append(random.randint(1, 1000))
print item
It will print numbers non-stop until memory is used up, or killed by system/user.
Understand the internal reason, let's discuss the solutions. Here are a few:
1. make a copy of origin list
Iterating the origin list, and modify the copied one.
result = l[:]
for item in l:
if somecond(item):
result.append(Obj())
2. control when the loop ends
Instead of handling control to python, you decides how to iterate the list:
length = len(l)
for index in range(length):
if somecond(l[index]):
l.append(Obj())
Before iterating, calculate the list length, and only loop length times.
3. store added objects in a new list
Instead of modifying the origin list, store new object in a new list and concatenate them afterward.
added = [Obj() for item in l if somecond(item)]
l.extend(added)
You can do this.
bonus_rows = []
for a in myarr:
if somecond(a):
bonus_rows.append(newObj())
myarr.extend( bonus_rows )
Access your list elements directly by i. Then you can append to your list:
for i in xrange(len(myarr)):
if somecond(a[i]):
myarr.append(newObj())
make copy of your original list, iterate over it,
see the modified code below
for a in myarr[:]:
if somecond(a):
myarr.append(newObj())
I had a similar problem today. I had a list of items that needed checking; if the objects passed the check, they were added to a result list. If they didn't pass, I changed them a bit and if they might still work (size > 0 after the change), I'd add them on to the back of the list for rechecking.
I went for a solution like
items = [...what I want to check...]
result = []
while items:
recheck_items = []
for item in items:
if check(item):
result.append(item)
else:
item = change(item) # Note that this always lowers the integer size(),
# so no danger of an infinite loop
if item.size() > 0:
recheck_items.append(item)
items = recheck_items # Let the loop restart with these, if any
My list is effectively a queue, should probably have used some sort of queue. But my lists are small (like 10 items) and this works too.
You can use an index and a while loop instead of a for loop if you want the loop to also loop over the elements that is added to the list during the loop:
i = 0
while i < len(myarr):
a = myarr[i];
i = i + 1;
if somecond(a):
myarr.append(newObj())
Expanding S.Lott's answer so that new items are processed as well:
todo = myarr
done = []
while todo:
added = []
for a in todo:
if somecond(a):
added.append(newObj())
done.extend(todo)
todo = added
The final list is in done.
Alternate solution :
reduce(lambda x,newObj : x +[newObj] if somecond else x,myarr,myarr)
Assuming you are adding at the last of this list arr, You can try this method I often use,
arr = [...The list I want to work with]
current_length = len(arr)
i = 0
while i < current_length:
current_element = arr[i]
do_something(arr[i])
# Time to insert
insert_count = 1 # How many Items you are adding add the last
arr.append(item_to_be inserted)
# IMPORTANT!!!! increase the current limit and indexer
i += 1
current_length += insert_count
This is just boilerplate and if you run this, your program will freeze because of infinite loop. DO NOT FORGET TO TERMINATE THE LOOP unless you need so.