I'm struggling to grasp the problem here. I already tried everything but the issue persist.
Basically I have a list of random numbers and when I try to compare the vaalues inside loop it throws "IndexError: list index out of range"
I even tried with range(len(who) and len(who) . Same thing. When put 0 instead "currentskill" which is int variable it works. What I don't understand is why comparing both values throws this Error. It just doesn't make sence...
Am I not comparing a value but the index itself ???
EDIT: I even tried with print(i) / print(who[i] to see if everything is clean and where it stops, and I'm definitelly not going outside of index
who = [2, 0, 1]
currentskill = 1
for i in who:
if who[i] == currentskill: # IndexError: list index out of range
who.pop(i)
The problem is once you start popping out elements list size varies
For eg take a list of size 6
But you iterate over all indices up to len(l)-1 = 6-1 = 5 and the index 5 does not exist in the list after removing elements in a previous iteration.
solution for this problem,
l = [x for x in l if x]
Here x is a condition you want to implement on the element of the list which you are iterating.
As stated by #Hemesh
The problem is once you start popping out elements list size varies
Problem solved. I'm just popping the element outside the loop now and it works:
def deleteskill(who, currentskill):
temp = 0
for i in range(len(who)):
if who[i] == currentskill:
temp = i
who.pop(temp)
There are two problems in your code:
mixing up the values and indexes, as pointed out by another answer; and
modifying the list while iterating over it
The solution depends on whether you want to remove just one item, or potentially multiple.
For removing just one item:
for idx, i in enumerate(who)::
if i == currentskill:
who.pop(idx)
break
For removing multiple items:
to_remove = []
for idx, i in enumerate(who)::
if i == currentskill:
to_remove.append[idx]
for idx in reversed(to_remove):
who.pop(idx)
Depending on the situation, it may be easier to create a new list instead:
who = [i for i in who if i != currentskill]
Your logic is wrong. To get the index as well as the value, use the built-in function enumerate:
idx_to_remove = []
for idx, i in enumerate(who)::
if i == currentskill:
idx_to_remove.append[idx]
for idx in reversed(idx_to_remove):
who.pop(idx)
Edited after suggestion from #sabik
Related
I apologize if this is a duplicate, I tried my best to find an existing question but was unsuccessful.
Recently, I've run into a couple of problems where I've needed to find the element in a list that produces the max/min value when a calculation is performed. For example, a list of real numbers where you want to find out which element produces the highest value when squared. The actual value of the squared number is unimportant, I just need the element(s) from the list that produces it.
I know I can solve the problem by finding the max, then making a pass through the list to find out which values' square matches the max I found:
l = [-0.25, 21.4, -7, 0.99, -21.4]
max_squared = max(i**2 for i in l)
result = [i for i in l if i**2 == max_squared]
but I feel like there should be a better way to do it. Is there a more concise/one-step solution to this?
This will return you just the element which gives the max when squared.
result = max(l, key = lambda k: k**2)
It does not get much better if you need the value in a list f.e. to see how often it occures. You can remeber the source element as well if you do not need that:
l = [-0.25, 21.4, -7, 0.99, -21.4]
max_squared = max( (i**2, i) for i in l) # remeber a tuple, with the result coming first
print(max_squared[1]) # print the source number (2nd element of the tuple)
Output:
21.4
Your calculation does only return the first occurence of abs(24.1) because max only returns one value, not two - if you need both, you still need to do:
print( [k for k in l if abs(k) == max_squared[1]])
to get
[21.4,-21.4]
I am trying to code up a numerical clustering tool. Basically, I have a list (here called 'product') that should be transformed from an ascending list to a list that indicates linkage between numbers in the data set. Reading in the data set, removing carriage returns and hyphens works okay, but manipulating the list based on the data set is giving me a problem.
# opening file and returning raw data
file = input('Data file: ')
with open(file) as t:
nums = t.readlines()
t.close()
print(f'Raw data: {nums}')
# counting pairs in raw data
count = 0
for i in nums:
count += 1
print(f'Count of number pairs: {count}')
# removing carriage returns and hyphens
one = []
for i in nums:
one.append(i.rsplit())
new = []
for i in one:
for a in i:
new.append(a.split('-'))
print(f'Data sets: {new}')
# finding the range of the final list
my_list = []
for i in new:
for e in i:
my_list.append(int(e))
ran = max(my_list) + 1
print(f'Range of final list: {ran}')
# setting up the product list
rcount = count-1
product = list(range(ran))
print(f'Unchanged product: {product}')
for i in product:
for e in range(rcount):
if product[int(new[e][0])] < product[int(new[e][1])]:
product[int(new[e][1])] = product[int(new[e][0])]
else:
product[int(new[e][0])] = product[int(new[e][1])]
print(f'Resulting product: {product}')
I expect the result to be [0, 1, 1, 1, 1, 5, 5, 7, 7, 9, 1, 5, 5], but am met with a 'list index out of range' when using a different data set.
the data set used to give the above desired product is as follows: '1-2\n', '2-3\n', '3-4\n', '5-6\n', '7-8\n', '2-10\n', '11-12\n', '5-12\n', '\n'
However, the biggest issue I am facing is using other data sets. If there is not an additional carriage return, as it turns out, I will have the list index out of range error.
I can't quite figure out what you're actually trying to do here. What does "indicates linkages" mean, and how does the final output do so? Also, can you show an example of a dataset where it actually fails? And provide the actual exception that you get?
Regardless, your code is massively over-complicated, and cleaning it up a little may also fix your index issue. Using nums as from your sample above:
# Drop empty elements, split on hyphen, and convert to integers
pairs = [list(map(int, item.split('-'))) for item in nums if item.strip()]
# You don't need a for loop to count a list
count = len(pairs)
# You can get the maximum element with a nested generator expression
largest = max(item for p in pairs for item in p)
Also, in your final loop you're iterating over product while also modifying it in-place, which tends to not be a good idea. If I had more understanding of what you're trying to achieve I might be able to suggest a better approach.
I am doing this problem a friend gave me where you are given 2 arrays say (a[1,2,3,4] and b[8,7,9,2,1]) and you have to find not common elements.
Expected output is [3,4,8,7,9]. Code below.
def disjoint(e,f):
c = e[:]
d = f[:]
for i in range(len(e)):
for j in range(len(f)):
if e[i] == f[j]:
c.remove(e[i])
d.remove(d[j])
final = c + d
print(final)
print(disjoint(a,b))
I tried with nested loops and creating copies of given arrays to modify them then add them but...
def disjoint(e,f):
c = e[:] # list copies
d = f[:]
for i in range(len(e)):
for j in range(len(f)):
if e[i] == f[j]:
c.remove(c[i]) # edited this line
d.remove(d[j])
final = c + d
print(final)
print(disjoint(a,b))
when I try removing common element from list copies, I get different output [2,4,8,7,9]. why ??
This is my first question in this website. I'll be thankful if anyone can clear my doubts.
Using sets you can do:
a = [1,2,3,4]
b = [8,7,9,2,1]
diff = (set(a) | set(b)) - (set(a) & set(b))
(set(a) | set(b)) is the union, set(a) & set(b) is the intersection and finally you do the difference between the two sets using -.
Your bug comes when you remove the elements in the lines c.remove(c[i]) and d.remove(d[j]). Indeed, the common elements are e[i]and f[j] while c and d are the lists you are updating.
To fix your bug you only need to change these lines to c.remove(e[i]) and d.remove(f[j]).
Note also that your method to delete items in both lists will not work if a list may contain duplicates.
Consider for instance the case a = [1,1,2,3,4] and b = [8,7,9,2,1].
You can simplify your code to make it works:
def disjoint(e,f):
c = e.copy() # [:] works also, but I think this is clearer
d = f.copy()
for i in e: # no need for index. just walk each items in the array
for j in f:
if i == j: # if there is a match, remove the match.
c.remove(i)
d.remove(j)
return c + d
print(disjoint([1,2,3,4],[8,7,9,2,1]))
Try it online!
There are a lot of more effecient way to achieve this. Check this stack overflow question to discover them: Get difference between two lists. My favorite way is to use set (like in #newbie's answer). What is a set? Lets check the documentation:
A set object is an unordered collection of distinct hashable objects. Common uses include membership testing, removing duplicates from a sequence, and computing mathematical operations such as intersection, union, difference, and symmetric difference. (For other containers see the built-in dict, list, and tuple classes, and the collections module.)
emphasis mine
Symmetric difference is perfect for our need!
Returns a new set with elements in either the set or the specified iterable but not both.
Ok here how to use it in your case:
def disjoint(e,f):
return list(set(e).symmetric_difference(set(f)))
print(disjoint([1,2,3,4],[8,7,9,2,1]))
Try it online!
i am trying to delete zero values from the list using the below code
for id,row in enumerate(list):
if row[0]=="0":
del list(id)
this works fine for input like
[0,1,3,0,9,10,0,3,0,6]
but doesn't work as expected for inputs like
[0,0,1,3,4,0,0,4,5,6,0,0].
output: [0,1,3,4,0,4,5,6,0]
I guess its because the element right after the deleted one gets the id of the deleted element and enumerate increments the id which leaves the element after the one which is deleted unchecked.
so what can be done to check all the elements ? is there a better way ?
I made a little change to your code:
mylist = [0,0,1,3,4,0,0,4,5,6,0,0]
for id,row in reversed(list(enumerate(mylist))):
if(row==0):
del mylist[id]
If you loop your list in the way you did (from start to end) and delete an element while doing it, you'll end up jumping indexes because python does not recognize that an element has been deleted from the list.
If you have an array with 10 elements inside and you delete the first (idx 0), in the next iteration you will be at index 1, but the array has been modified, so your idx 1 is the idx 2 of your array before the deletion, and the real idx 1 will be skipped.
So you just need to loop your array in reverse mode, and you won't miss indexes.
If you print the value of mylist, you'll get [1, 3, 4, 4, 5, 6].
This problem is documented on this python page under 8.3:
https://docs.python.org/3/reference/compound_stmts.html
They suggest doing it this way by using a slice. It works for me:
a = [-2,-4,3,4]
for x in a[:]:
if x < 0: a.remove(x)
print ('contents of a now: ')
print(*a)
enumerate returns an object called enumerate object and it is iterable not actually a list. second thing row is not a list it is not subscriptable.
for i,row in enumerate(l):
if row==0:
del(l[i])
you will not get result you want this way.
try this:
t=[] #a temporary list
for i in l:
if i!=0:
t.append(i)
t will contain sublist of l with non zero elements.
put the above inside a function and return the list t .
I have a list of about 20-30 items [strings].
I'm able to print them out in my program just fine - but I'd like to save some space, and merge items that are shorter...
So basically, if I have 2 consecutive items that the combined lengths are less than 30, I want to join those to items as a single entry in the list - with a / between them
I'm not coming up with a simple way of doing this.
I don't care if I do it in the same list, or make a new list of items... it's all happening inside 1 function...
You need to loop through the list and keep joining items till they satisfy your requirement (size 30). Then add them to a new list when an element grows that big.
l=[] # your new list
buff=yourList[0] if len(yourList)>0 else "" # hold strings till they reach desired length
for i in range(1,len(yourList)):
# check if concatenating will exceed the size or not
t=yourList[i]
if (len(buff) + len(t) + 1) <= 30:
buff+="/"+t
else:
l.append(buff)
buff=t
l.append(buff) # since last element is yet to be inserted
You can extend method of list as follows:
a = [1,2,3]
b = [4,5,6]
a.append('/')
a.extend(b)
You just need to check the size of two list a and b as per your requirements.
I hope I understood your problem !
This code worked for me, you can check to see if that's what you wanted, it's a bit lenghty but it works.
list1 = yourListOfElements
for elem in list1:
try: # Needs try/except otherwise last iteration would throw an indexerror
listaAUX = [] # Auxiliar list to check length and join smaller elements. You can probably eliminate this using list slicing
listaAUX.append(elem)
listaAUX.append(list1[list1.index(elem)+1])
if len(listaAUX[0]) + len(listaAUX[1]) < 30:
concatenated = '/'.join(listaAUX)
print(concatenated)
else:
print(elem)
except IndexError:
print(elem)