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How to handle operating on items in a list without perfectly even len()?

I'm trying to operate on every 5 items in a list, but can't figure out how to handle the remaining items if they don't divide evenly into 5. Right now I'm using modulo, but I can't shake the feeling it's not quite the right answer. Here's an example...
list = ["ValA","ValB","ValC","ValD","ValE","ValF","ValG","ValH","ValI","ValJ","ValK","ValL","ValM","ValN",]
newlist = []
i = 0
for o in list:
i += 1
newlist.append(o)
if i % 5 == 0:
for obj in newlist:
function_for(obj)
newlist.clear()
This code will execute function_for() twice, but not a third time to handle the remaining 4 values. If I add an 'else' statement it runs on every execution.
What's the correct way to handle a situation like this?

This way is pretty easy, if you don't mind modifying the list:
mylist = ["ValA","ValB","ValC","ValD","ValE","ValF","ValG","ValH","ValI","ValJ","ValK","ValL","ValM","ValN",]
while mylist:
function_for( mylist[:5] )
mylist = mylist[5:]

You can also check if the index is equal to the length of the list. (Additionally, it is more idiomatic to use enumerate instead of a counter variable here.)
lst = ["ValA","ValB","ValC","ValD","ValE","ValF","ValG","ValH","ValI","ValJ","ValK","ValL","ValM","ValN",]
newlist = []
for i, o in enumerate(lst, 1):
newlist.append(o)
if i % 5 == 0 or i == len(lst):
print(newlist)
newlist.clear()

If the input number is in the list add its index to a new one

I want to check if the input number is in the list, and if so - add its index in the original list to the new one. If it's not in the list - I want to add a -1.
I tried using the for loop and adding it like that, but it is kind of bad on the speed of the program.
n = int(input())
k = [int(x) for x in input().split()]
z = []
m = int(input())
for i in range(m):
a = int(input())
if a in k: z.append(k.index(a))
else: z.append(-1)
The input should look like this :
3
2 1 3
1
8
3
And the output should be :
1
-1
2
How can I do what I'm trying to do more efficiently/quickly

There are many approaches to this problem. This is typical when you're first starting in programming as, the simpler the problem, the more options you have. Choosing which option depends what you have and what you want.
In this case we're expecting user input of this form:
3
2 1 3
1
8
3
One approach is to generate a dict to use for lookups instead of using list operations. Lookup in dict will give you better performance overall. You can use enumerate to give me both the index i and the value x from the list from user input. Then use int(x) as the key and associate it to the index.
The key should always be the data you have, and the value should always be the data you want. (We have a value, we want the index)
n = int(input())
k = {}
for i, x in enumerate(input().split()):
k[int(x)] = i
z = []
for i in range(n):
a = int(input())
if a in k:
z.append(k[a])
else:
z.append(-1)
print(z)
k looks like:
{2: 0, 1: 1, 3: 2}
This way you can call k[3] and it will give you 2 in O(1) or constant time.
(See. Python: List vs Dict for look up table)
There is a structure known as defaultdict which allows you to specify behaviour when a key is not present in the dictionary. This is particularly helpful in this case, as we can just request from the defaultdict and it will return the desired value either way.
from collections import defaultdict
n = int(input())
k = defaultdict(lambda: -1)
for i, x in enumerate(input().split()):
k[int(x)] = i
z = []
for i in range(n):
a = int(input())
z.append(k[a])
print(z)
While this does not speed up your program, it does make your second for loop easier to read. It also makes it easier to move into the comprehension in the next section.
(See. How does collections.defaultdict work?
With these things in place, we can use, yes, list comprehension, to very minimally speed up the construction of z and k. (See. Are list-comprehensions and functional functions faster than “for loops”?
from collections import defaultdict
n = int(input())
k = defaultdict(lambda: -1)
for i, x in enumerate(input().split()):
k[int(x)] = i
z = [k[int(input())] for i in range(n)]
print(z)
All code snippets print z as a list:
[1, -1, 2]
See Printing list elements on separated lines in Python if you'd like different print outs.
Note: The index function will find the index of the first occurrence of the value in a list. Because of the way the dict is built, the index of the last occurrence will be stored in k. If you need to mimic index exactly you should ensure that a later index does not overwrite a previous one.
for i, x in enumerate(input().split()):
x = int(x)
if x not in k:
k[x] = i

Adapt this solution for your problem.
def test(list1,value):
try:
return list1.index(value)
except ValueError as e:
print(e)
return -1
list1=[2, 1, 3]
in1 = [1,8,3]
res= [test(list1,i) for i in in1]
print(res)
output
8 is not in list
[1, -1, 2]

Nested For Loop Copies the last line in set of indicies

for each in Dti:
i = 0
for each in Dti[0]:
xbi[t][i] = Dti[t][i]
print(t)
i = i + 1
t = t + 1
this is just a test that I'm doing to figure out why my complicated code isn't working. I'm trying to iterate through a list and then each value in the list to set a new list of lists equal to that value. I know I could just set them equal to each other, but it needs to be done this way for my more complicated program. Any tips? I'm getting Dti[-1] for each xbi[t]. I've tried with while too and got the same results

Try something like this:
for t, D in enumerate(Dti)
for i, d in enumerate(D):
xbi[t][i] = d
print(t)

You can use slicing in assignments to replace one list's elements with the elements of another list:
for t, row in enumerate(Dti):
xbi[t][:] = row

How can I alter the index of loop for a list

I've a list and I want to display the content with index by manually changing the value of index. I've tried to do it in two ways but it is not working.
way 1:
x=['hello','how','are','you','hope','you','are','fine','I','am','doing','work','in','python']
for i,item in enumerate(x):
print (i,item)
if item == 'are':
i+=2
way 2:
x=['hello','how','are','you','hope','you','are','fine','I','am','doing','work','in','python']
for i in xrange(len(x)):
print x[i]
if item == 'are':
i+=2

The issue here is that both enumerate() and xrange() generate a series of values that is unaffected by external changes to i, which is an unconnected variable. Using #efferalgans's suggestion the best way might be to call the next() function twice, but this will mean storing a reference to the generator. So you want to do something like this:
x=['hello','how','are','you','hope','you','are','fine','I','am','doing','work','in','python']
g = enumerate(x)
for i,item in g:
print (i,item)
if item == 'are':
next(g)
next(g)
from which I see the result
0 hello
1 how
2 are
5 you
6 are
9 am
10 doing
11 work
12 in
13 python

Just use a while loop. It's the simplest, clearest and most flexible solution:
i = 0
while i < len(x):
print x[i]
if x[i] == 'are':
i+=1
i += 1
As you can see, looping over a string is simple and you are completely free to change the index on the fly -- i is just a normal variable. It's not clear what your code was actually meant to do, so the above does what your code would (presumably) have done if it worked: It will skip the word that follows the word "are".

Memoryerror with too big list

I'm writing script in python, and now I have to create pretty big list exactly containing 248956422 integers. The point is, that some of this "0" in this table will be changed for 1,2 or 3, cause I have 8 lists, 4 with beginning positions of genes, and 4 with endings of them.
The point is i have to iterate "anno" several time cause numbers replacing 0 can change with other iteration.
"Anno" has to be written to the file to create annotation file.
Here's my question, how can I divide, or do it on-the-fly , not to get memoryerror including replacing "0" for others, and 1,2,3s for others.
Mabye rewriting the file? I'm waitin for your advice, please ask me if it is not so clear what i wrote :P .
whole_st_gen = [] #to make these lists more clear for example
whole_end_gen = [] # whole_st_gen has element "177"
whole_st_ex = [] # and whole_end_gen has "200" so from position 177to200
whole_end_ex = [] # i need to put "1"
whole_st_mr = [] # of course these list can have even 1kk+ elements
whole_end_mr = [] # note that every st/end of same kind have equal length
whole_st_nc = []
whole_end_nc = [] #these lists are including some values of course
length = 248956422
anno = ['0' for i in range(0,length)] # here i get the memoryerror
#then i wanted to do something like..
for j in range(0, len(whole_st_gen)):
for y in range(whole_st_gen[j],whole_end_gen[j]):
anno[y]='1'

You might be better of by determine the value of each element in anno on the fly:
def anno():
for idx in xrange(248956422):
elm = "0"
for j in range(0, len(whole_st_gen)):
if whole_st_gen[j] <= idx < whole_end_gen[j]:
elm = "1"
for j in range(0, len(whole_st_ex)):
if whole_st_ex[j] <= idx < whole_end_ex[j]:
elm = "2"
for j in range(0, len(whole_st_mr)):
if whole_st_mr[j] <= idx < whole_end_mr[j]:
elm = "3"
for j in range(0, len(whole_st_nc)):
if whole_st_nc[j] <= idx < whole_end_nc[j]:
elm = "4"
yield elm
Then you just iterate using for elm in anno().
I got an edit proposal from the OP suggesting one function for each of whole_*_gen, whole_st_ex and so on, something like this:
def anno_st():
for idx in xrange(248956422):
elm = "0"
for j in range(0, len(whole_st_gen)):
if whole_st_ex[j] <= idx <= whole_end_ex[j]:
elm = "2"
yield elm
That's of course doable, but it will only result in the changes from whole_*_ex applied and one would need to combine them afterwards when writing to file which may be a bit awkward:
for a, b, c, d in zip(anno_st(), anno_ex(), anno_mr(), anno_nc()):
if d != "0":
write_to_file(d)
elif c != "0":
write_to_file(c)
elif b != "0":
write_to_file(b)
else:
write_to_file(a)
However if you only want to apply some of the change sets you could write a function that takes them as parameters:
def anno(*args):
for idx in xrange(248956422):
elm = "0"
for st, end, tag in args:
for j in range(0, len(st)):
if st <= idx < end[j]:
elm = tag
yield tag
And then call by supplying the lists (for example with only the two first changes):
for tag in anno((whole_st_gen, whole_end_gen, "1"),
(whole_st_ex, whole_end_ex, "2")):
write_to_file(tag)

You could use a bytearray object to have a much more compact memory representation than a list of integers:
anno = bytearray(b'\0' * 248956422)
print(anno[0]) # → 0
anno[0] = 2
print(anno[0]) # → 2
print(anno.__sizeof__()) # → 248956447 (on my computer)

Instead of creating a list using list comprehension I suggest to create an iterator using a generator-expression which produce the numbers on demand instead of saving all of them in memory.Also you don't need to use the i in your loop since it's just a throw away variable which you don't use it.
anno = ('0' for _ in range(0,length)) # In python 2.X use xrange() instead of range()
But note that and iterator is a one shot iterable and you can not use it after iterating over it one time.If you want to use it for multiple times you can create N independent iterators from it using itertools.tee().
Also note that you can not change it in-place if you want to change some elements based on a condition you can create a new iterator by iterating over your iterator and applying the condition using a generator expression.
For example :
new_anno =("""do something with i""" for i in anno if #some condition)