def rev(n):
for i in range(int(len(n)//2)):
temp = n[i]
n[i] = n[len(n)-i-1]
n[len(n)-i-1] = temp
or,n[i], n[len(n)-i-1] = n[len(n)-i-1], n[i]
return n
n=[34,45,56,67]
print(rev(n))
Above code doesn't reverse the list even the logic is correct still the output is same as input.
Can anyone help me with that as i am little bit confused.
The intention appears to be to reverse the contents of a list in situ.
The most common way to do this is to create a new list with:
mylist[::-1]
...then... copy over the original like this:
mylist[:] = mylist[::-1]
However, the intent appears to be to use a custom loop to reverse in place.
def rev(_list):
i, j = 0, len(_list)
while (j := j - 1) > i:
_list[i], _list[j] = _list[j], _list[i]
i += 1
return _list
This demonstrates the correct element swap syntax and also obviates the need to create a new list. It is however very slow in comparison to traditional techniques
Related
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l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.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.
Let's say I have a string
S = "qwertyu"
And I want to build a list using recursion so the list looks like
L = [u, y, t, r, e, w, q]
I tried to write code like this:
def rec (S):
if len(S) > 0:
return [S[-1]].append(rec(S[0:-1]))
Ideally I want to append the last element of a shrinking string until it reaches 0
but all I got as an output is None
I know I'm not doing it right, and I have absolutely no idea what to return when the length of S reaches 0, please show me how I can make this work
(sorry the answer has to use recursion, otherwise it won't bother me)
Thank you very much!!!
There are many simpler ways than using recursion, but here's one recursive way to do it:
def rec (S):
if not S:
return []
else:
temp = list(S[-1])
temp.extend(rec(S[:-1]))
return temp
EDIT:
Notice that the base case ensures that function also works with an empty string. I had to use temp, because you cannot return list(S[-1]).extend(rec(S[:-1])) due to it being a NoneType (it's a method call rather than an object). For the same reason you cannot assign to a variable (hence the two separate lines with temp). A workaround would be to use + to concatenate the two lists, like suggested in Aryerez's answer (however, I'd suggest against his advice to try to impress people with convoluted one liners):
def rec (S):
if not S:
return []
else:
return list(S[-1]) + rec(S[:-1])
In fact using + could be more efficient (although the improvement would most likely be negligible), see answers to this SO question for more details.
This is the simplest solution:
def rec(S):
if len(S) == 1:
return S
return S[-1] + rec(S[:-1])
Or in one-line, if you really want to impress someone :)
def rec(S):
return S if len(S) == 1 else S[-1] + rec(S[:-1])
Since append mutates the list, this is a bit difficult to express recursively. One way you could do this is by using a separate inner function that passes on the current L to the next recursive call.
def rec(S):
def go(S, L):
if len(S) > 0:
L.append(S[-1])
return go(S[0:-1], L)
else:
return L
return go(S, [])
L = [i for i in S[::-1]]
It should work.
Hello guys I am trying to implement insertion recursively but my list is not getting sorted. Any help or improvements are welcome.
sequence = [1,4,3,5,7,6,8,2]
def insertion(seq):
for i in range(len(seq)):
key = seq[i]
j = i
print(seq)
while j > 0 and seq[j-1] > key:
j = j - 1
seq[j] == key
insertion(sequence)
You have several errors in your code:
Your indentation is incorrect, fix that first.
You're missing a statement that should come as the first in the body
of the while loop to assign seq[j] to be the element that follows
it.
The seq[j] == key statement is a no-op -- this is not a ==
situation but rather a = one.
Your implementation isn't recursive, so once you get this code working, start over.
I found this code online.
def quick_sort(items):
""" Implementation of quick sort """
if len(items) > 1:
pivot_index = len(items) / 2
smaller_items = []
larger_items = []
for i, val in enumerate(items):
if i != pivot_index:
if val < items[pivot_index]:
smaller_items.append(val)
else:
larger_items.append(val)
quick_sort(smaller_items)
quick_sort(larger_items)
items[:] = smaller_items + [items[pivot_index]] + larger_items
The one line that gives me trouble is the last one. I believe it's basic concatenation, however, when I change "items[:]" to "items", the algorithm fails. What is special about the [:] at the end of the list?
If anyone one can help, I would really appreciate. Thank you in advance!
This is in-place assignment, without replacing the list object with a new object.
This way, the output is placed in the original list and can be read by the caller.
I am new to programming, and I am trying to write a Vigenère Encryption Cipher using python. The idea is very simple and so is my function, however in this line:
( if((BinKey[i] == 'b')or(BinKey[i+1] == 'b')): )
It seems that I have an index problem, and I can't figure out how to fix it.
The error message is:
IndexError: string index out of range
I tried to replace the i+1 index by another variable equal to i+1, since I thought that maybe python is re-incrementing the i, but it still won't work.
So my questions are:
How to fix the problem, and what have I done wrong?
Looking at my code, what can I learn to improve my programming skills?
I want to build a simple interface to my program (which will contain all the encryption ciphers), and all I came up with from Google is pyqt, but it just seems too much work for a very simple interface, so is there a simpler way to build an interface? (I am working with Eclipse Indigo and pydev with Python3.x)
The Vigenère Encryption function (which contains the line that causes the problem) is:
def Viegner_Encyption_Cipher(Key,String):
EncryptedMessage = ""
i = 0
j = 0
BinKey = Bin_It(Key)
BinString = Bin_It(String)
BinKeyLengh = len(BinKey)
BinStringLengh = len(BinString)
while ((BinKeyLengh > i) and (BinStringLengh > j)):
if((BinKey[i] == 'b')or(BinKey[i+1] == 'b')):
EncryptedMessage = EncryptedMessage + BinKey[i]
else:
EncryptedMessage = EncryptedMessage + Xor(BinKey[i],BinString[j])
i = i + 1
j = j + 1
if (i == BinKeyLengh):
i = i+j
return EncryptedMessage
This is the Bin_It function:
def Bin_It(String):
TheBin = ""
for Charactere in String:
TheBin = TheBin + bin(ord(Charactere))
return TheBin
And finally this is the Xor function:
def Xor(a,b):
xor = (int(a) and not int(b)) or (not int(a) and int(b))
if xor:
return chr(1)
else:
return chr(0)
In your while condition, you are ensuring that i < len(BinKey). This means that BinKey[i] will be valid, but BinKey[i+1] will not be valid at the last iteration of your loop, as you will then be accessing BinKey[len(BinKey)], which is one past the end of your string. Strings in python start at 0 and end at len-1 inclusive.
To avoid this, you can update your loop criterion to be
while BinKeyLength > i+1 and ...:
You can either change
while ((BinKeyLengh > i) and (BinStringLengh > j)):
to
while ((BinKeyLengh > i-1) and (BinStringLengh > j)):
or change
if((BinKey[i] == 'b')or(BinKey[i+1] == 'b')):
to
if((BinKey[i] == 'b') or (BinKeyLengh > i-1 and BinKey[i+1] == 'b')):
This will avoid trying to go into BinKey[BinKeyLength], which is out of scope.
Looking at my code, what can I learn to improve my programming skills?
Looping over an index is not idiomatic Python. It's better to loop over the elements of an iterator where possible. After all, that's usually what you're interested in: for i in... is often followed by my_list[i].
In this example, you should use the built-in function zip (or itertools.izip if your code is lazy, though this isn't necessary in Python 3), which gives you pairs of values from two or more iterators, and stops when the shortest one is exhausted.
for key_char, string_char in zip(BinKey, BinString): # takes values sequentially from
# BinKey and BinString
# and binds them to the names
# key_char and string_char
# do processing on key_char and string_char
If you really must do a while loop on an index, then put the test the other way around so it's clearer what you're doing. Compare
while len(BinKey) > i and len(BinString) > j: # this looks like len(BinKey) and
# len(BinString) are varying and you're
# comparing them to static variables i and j
with
while i < len(BinKey) and j < len(BinString): # this looks like you're varying i and j
# and comparing them to len(BinKey) and len(BinString)
Which better communicates the purpose of the loop?
Finally, the clause
if (i == BinKeyLengh):
i = i+j
doesn't seem to do anything. If i == BinKeyLength then the while loop will stop in a moment anyway.
I think your error, as Python interpreter says, is that you accessing invalid array positions.
In order to solve this, unlike it's being said, you should change your code to
while (BinKeyLength > i+2 and ...):
This is because in the last step, BinKeyLength = i+2, then i+1 is BinKeyLength-1, which is the last position of your array.
Concerning your programming skills I recommend you two things:
Be the code. Sounds mystic but the most important thing that is missing here is figuring out which indices numbers are used.
As it has been said by rubik, follow some style guides like PEP8 style guide.