How do I check the index of a an element in a list? (Python)

Question

list = [('ba',4), ('hh',5), ('gg', 25)]

How do I do:

list.index('hh') ...and returns 1?

Then, how do I sort it by the 25, 5, 4?

What if I have 2 lists:

list1 = [('ba',4), ('hh',5), ('gg', 25)]
list2 = [('ja',40), ('hgh',88), ('hh', 2)]

how do I do a for each?

for item in l1:
    if item[0] in l2[0 of the tuple]:

Answer 1

For the finding

>>> L = [('ba',4), ('hh',5), ('gg', 25)]
>>> [ i for i,l in enumerate(L) if l[0] == 'hh' ][0]
1

You need to decide what to do if it is found multiple times or not at all - the above will throw IndexError if not found and return the first if it is found multiple times.

For the sorting

>>> L = [('ba',4), ('hh',5), ('gg', 25)]
>>> sorted(L, key=lambda x: x[1])
[('ba', 4), ('hh', 5), ('gg', 25)]

Answer 2

to sort the list u can use a custom sort method some thing like this

x = [('ba',4), ('hh',5), ('gg', 25)]

def sortMethod(x,y):
    if x[1] < y[1]:return 1
    elif x[1] > y[1]:return -1
    else: return 0


print x         #unsorted
x.sort(sortMethod)
print x         #sorted

Answer 3

First of, don't use list as the name for a variable, as it shadows the built-in list function.

1. You can use enumerate to pair up list elements and their index:

   >>> l = [('ba',4), ('hh',5), ('gg', 25)]
   >>> [i for i, e in enumerate(l) if e[0] == 'hh']
   [1]
   

2. For sorting you can use a lambda expression as shown by others, or you can pass an operator.itemgetter as the key argument to sorted:

   >>> from operator import itemgetter
   >>> sorted(l, key=itemgetter(1))
   [('ba', 4), ('hh', 5), ('gg', 25)]
   

3. In-place sorting is also possible, using the sort method on lists:

   >>> l.sort(key=itemgetter(1))
   

Answer 4

you can also have your list in dictionary form

list1 = [('ba',4), ('hh',5), ('gg', 25)]
dict1 = dict(list1)

print dict1['hh']
5

dicts are faster then list if you need to search like that.

btw, overriding built-in type list to variables are not good idea list = [('ba',4), ('hh',5), ('gg', 25)].

Answer 5

For the sort, you should use itemgetter

>>> import operator
>>> L = [('ba',4), ('hh',5), ('gg', 25)]
>>> sorted(L, key=operator.itemgetter(1))
[('ba', 4), ('hh', 5), ('gg', 25)]

Answer 6

I think Nick's sorting answer is good, but his find method unnecessarily iterates over the entire list, even after it has found a match. With a small change it can be fixed to stop iterating as soon as it finds the first element:

index = (i for i,l in enumerate(l) if l[0] == 'aa').next()

Or in Python 3:

index = next(i for i,l in enumerate(l) if l[0] == 'aa')

Answer 7

For the last question, convert list2 into a set:

>>> list1 = [('ba',4), ('hh',5), ('gg', 25)]
>>> list2 = [('ja',40), ('hgh',88), ('hh', 2)]
>>> 
>>> wanted = set(a for (a,b) in list2)
>>> for x in list1:
...     if x[0] in wanted:
...         print x
... 
('hh', 5)

Answer 8

from itertools import imap

def find(iterable, item, key=None):
    """Find `item` in `iterable`.

    Return index of the found item or ``-1`` if there is none.

    Apply `key` function to items before comparison with
    `item`. ``key=None`` means an identity function.
    """
    it = iter(iterable) if key is None else imap(key, iterable)
    for i, e in enumerate(it):
        if e == item:
            return i
    return -1

Example:

L = [('ba', 4), ('hh', 5), ('gg', 25)]
print find(L, 'hh', key=lambda x: x[0])

Output:

1