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Python Check if a 2-dimensional list contains numbers of another list

I want to know if a list contains numbers from a list of a list, but nothing I tried worked as I wanted. For example if a list of a list in list2 = [[1,2,3],[4,5,6]] is in list1 = [4, 5, 6, 7] it should print out True, because the numbers of list1 are a whole list from list2. So because list1 contains the numbers 4, 5, 6 the output should be True.
Here is what I tried
for i in range(len(list2)):
if list1 in list2[i]:
print('True')
else:
print('False')
It prints out False two times (but I need it to print out True) and I get why
[1, 2, 3] in [1, 2, 3, 4]
>>> False
Is there a way to avoid this?
Or should I just use strings since it works with them just fine?

Finding if a collection of items is a subset of another is more efficiently handled by a set than a list. Casting your list to a set allow quick lookup with the set.issuperset method.
list_of_lists = [[1,2,3],[4,5,6]]
list1 = [4, 5, 6, 7]
list1_as_set = set(list1)
result = any(list1_as_set.issuperset(l) for l in list_of_lists) # True
The above code using any is equivalent to this solution using a for-loop.
for l in list_of_lists:
if list1_as_set.issuperset(l):
result = True
break
else:
result = False

Python Removing a sublist in a list without changing the index of other sublists in the list

Say for example:
list = { [1,2,3],[4,5,6],[3,4],[2,7,8,9] }
is there a way in python to remove a sublist and make sure that other sublists index's remain the same. So for example if i was to remove the sublist [3,4], could i make sure that the index of [2,7,8,9] remains as 3 in this case? if this is possible it would really be helpful! thanks!

Maybe. It depends on how you use the list and just how "listy" the resulting object needs to be.
l = [ [1,2,3],[4,5,6],[3,4],[2,7,8,9] ]
You could replace the sublist with something else like a None. Then your code would have to know to ignore None when it processes the list.
print(l[3])
l[2] = None
print(l[3])
or you could convert the list to a dict and delete the member. You can still index the object but since its now a dict, your code will have to treat it like a dict.
l = dict(enumerate(l))
print l[3]
del l[2]
print l[3]
These tricks will only work in some specialized environments.

You can just delete the elements in [3,4] and keep the empty sub-list.
>>> lst = [[1,2,3],[4,5,6],[3,4],[2,7,8,9]]
>>> del lst[2][:]
>>> lst
[[1, 2, 3], [4, 5, 6], [], [2, 7, 8, 9]]
Please note that you should not use list as a variable name as list is a built in function

But you can store the index with the value as a tuple. First make a modified list which has both index and value. Then you are free to remove any of the element.
lst = [[1,2,3],[4,5,6],[3,4],[2,7,8,9]]
modified = list(enumerate(lst))
To explain a bit:
modified=[]
for i,v in enumerate(lst):
modified.append((i,v))
print modified
Output:
[(0, [1, 2, 3]), (1, [4, 5, 6]), (2, [3, 4]), (3, [2, 7, 8, 9])]

How to pick the smallest value in a list when an iterative process is applied to the list

I need to find the smallest value in a series of lists. I understand the code for the smallest value in just one list:
>>> x = [1, 2, 3, 4, 5]
>>> print (min(x))
1
Simple enough. However, I would like to know if there is a way to write code that finds the smallest value for each list I have without stopping and adjusting the code (whether by an iterative process or some other means). Any help would be greatly appreciated. Thanks in advance!

First, make a list of lists out of your separate lists. For example, if you have lists A, B and C, the list of lists would be [A,B,C].
Now, to get a list of all the minimum values for each list in a list of lists lst:
[min(x) for x in lst]
To get the global minimum:
min(x for sublist in lst for x in sublist)
Demo:
>>> lst
[[1, 2, 3], [4, 5, 6], [7, 8, 9]]
>>> [min(x) for x in lst]
[1, 4, 7]
>>> min(x for sublist in lst for x in sublist)
1
Edit in reaction to OP's comment:
I just want the minimum value for each list. I don't need to compile all of the minimum values together into a new list
If you just want to print the minimum values for each list A, B, C, ... , you can do:
for lst in (A,B,C): # put as many lists as you like into the parentheses
print(min(lst))
Edit in reaction to
I am only accessing one list (in this case, the values are well depths) at a time. [..] What I would like to know is how to write a code that finds the smallest value in a list given that the iterator essentially changes the values in said list.
Just print(min(lst)) each time lst has changed.

Assuming you've got a list of lists, this should do it:
minimums = [min(l) for l in lists]

Concatenate List Object Name with a Number and Retain the List Python

I'm using python 2.7 I'm trying to figure out a way to change the names of my lists automatically.
Let me explain i have multiple lists
list1 = [1, 2, 3, 4, 5]
list2 = [4, 5, 9, 3]
list3 = [8, 4, 3, 2, 1]
I would like to call the lists in a loop to determine which lists contain or do not contain a particular number.
My first thought was
x = "list" + str(i) # (where i iterates in the loop)
print x
However, using the above code only gave me the string "list1"(when i=1).
What I want is to be able to call the list that is named list1 and use the .count() operator to determine whether or not the number exists if it doesn't i want to call the next list until I'm out of lists(there will eventually be up to 30 lists).
Thanks,
Ryan

You shouldn't approach it like this. Put your lists in a container to iterate over them instead:
In [5]: for l in (list1, list2, list3):
...: print l.count(2)
...:
1
0
1
What you could do in a real-life use case is create a list of lists and fill it dynamically.
Then to get the first list that contains a given number, you could do:
In [6]: lists = [list1, list2, list3]
In [7]: next(l for l in lists if 9 in l)
Out[7]: [4, 5, 9, 3]

put the list in dict:
list1 = [1,2.4]
list2 = [2,5,6]
dlist = {1:list1,2:list2}
for k in dlist:
print dlist[k]

iterating through a list removing items, some items are not removed

I'm trying to transfer the contents of one list to another, but it's not working and I don't know why not. My code looks like this:
list1 = [1, 2, 3, 4, 5, 6]
list2 = []
for item in list1:
list2.append(item)
list1.remove(item)
But if I run it my output looks like this:
>>> list1
[2, 4, 6]
>>> list2
[1, 3, 5]
My question is threefold, I guess: Why is this happening, how do I make it work, and am I overlooking an incredibly simple solution like a 'move' statement or something?

The reason is that you're (appending and) removing from the first list whereby it gets smaller. So the iterator stops before the whole list could be walked through.
To achieve what you want, do this:
list1 = [1, 2, 3, 4, 5, 6]
list2 = []
# You couldn't just make 'list1_copy = list1',
# because this would just copy (share) the reference.
# (i.e. when you change list1_copy, list1 will also change)
# this will make a (new) copy of list1
# so you can happily iterate over it ( without anything getting lost :)
list1_copy = list1[:]
for item in list1_copy:
list2.append(item)
list1.remove(item)
The list1[start:end:step] is the slicing syntax: when you leave start empty it defaults to 0, when you leave end empty it is the highest possible value. So list1[:] means everything in it. (thanks to Wallacoloo)
Like some dudes said, you could also use the extend-method of the list-object to just copy the one list to another, if this was your intention. (But I choosed the way above, because this is near to your approach.)
As you are new to python, I have something for you: Dive Into Python 3 - it's free and easy. - Have fun!

You're deleting items from list1 while you're iterating over it.
That's asking for trouble.
Try this:
>>> list1 = [1,2,3,4,5,6]
>>> list2 = []
>>> list2 = list1[:] # we copy every element from list1 using a slice
>>> del list1[:] # we delete every element from list1

Essential debugging skill: Add print statements. (or print functions in Python 3)
>>> list1= [1, 2, 3, 4, 5, 6]
>>> for item in list1:
... print item
... list1.remove(item)
... print list1
...
1
[2, 3, 4, 5, 6]
3
[2, 4, 5, 6]
5
[2, 4, 6]
Notice that Python is trying to step through the positions of the list, but you keep removing items from the list, making the positions become meaningless.
Python picks the item at position 0 from the list.
You then remove the item, changing the list.
Python then picks the item at position 1 from the list (appearing to skip an item)
You then remove that item, changing the list.
Python then picks the item at position 2 from the list (appearing to skip an item)
You then remove the item, changing the list.
Python would then like to pick the item at position 3, but there's no such item. So the loop stops.

You shouldn't modify a list while you iterate over it. This causes the iterator to point at the wrong item. After the first item is handled, the iterator is pointing at index = 1, but because you've removed an item, the next item is now at index zero, so it will be skipped. This is why you are only handling every other item.
Try:
list2.extend(list1) # This appends all items from list1 to list2.
del list1[:] # From ChristopheD's post.

Exactly as ChristopheD says.
Could do this:
list1 = [1, 2, 3, 4, 5, 6]
list2 = []
for item in list1:
list2.append(item)
list1 = []
That'll clear list1.
Edit He/she has updated their post. I'll leave this up as a slight alternative.

Try this instead:
list1 = [1, 2, 3, 4, 5, 6]
list2 = []
list2.extend(list1)
list1[:] = []

There is also the humble copy function (or deepcopy if you have complex objects and not just integers in your list):
from copy import copy
list2 = copy(list1)
You might get a more appropriate answer if you explain what you're trying to accomplish (unless you're just learning about python lists).
"Variables" in Python are just names/references, so the destructive copy you seem to want to do seems kind of strange. If you want list2 to have the same values as list1 you can just do:
list2 = list1 # now they are both referring to the same list
And if after that you want to use list1 for something else, you can just do:
list1 = ['A', 'B', 'C']

Using a list comprehension:
list2 = [item for item in list1]
Bind the name list2 to the the same object as list1:
list2 = list1
(Note that if you modify the contents of list1, list2 will be change accordingly.)
Create a copy of list1 and bind it to the name list2:
list2 = list1[:]
In this case list1 and list2 are different objects.

As you can see in other answers, you are trying to modify the list while iterating over it. This doesn't work. There are many ways to copy one list to another. I did some tests to see how fast each approach is:
>>> timeit.Timer('list2 = list1[:]', 'list1 = range(10**3)').timeit(10**6)
3.9134418964385986
>>> timeit.Timer('list2 = []; list2.extend(list1)', 'list1 = range(10**3)').timeit(10**6)
4.9082601070404053
>>> timeit.Timer('list2 = copy.copy(list1)', 'import copy; list1 = range(10**3)').timeit(10**6)
7.5023419857025146
>>> timeit.Timer('list2 = [i for i in list1]', 'list1 = range(10**3)').timeit(10**6)
95.697894811630249
The slice syntax is the fastest. It's much faster than using a list comprehension.
To clear a list, you can use:
del list1[:]

#potatocubed: Many of the answers given solve the trivial example you gave ("move list1 to list2"), but don't really explain the "why" of the deeper problem, which is modifying a list as you iterate over it. Study S.Lott's answer...