Find an element in a list of tuples

Read up on List Comprehensions

[ (x,y) for x, y in a if x  == 1 ]

Also read up up generator functions and the yield statement.

def filter_value( someList, value ):
    for x, y in someList:
        if x == value :
            yield x,y

result= list( filter_value( a, 1 ) )

for item in a:
   if 1 in item:
       print item

[tup for tup in a if tup[0] == 1]

There is actually a clever way to do this that is useful for any list of tuples where the size of each tuple is 2: you can convert your list into a single dictionary.

For example,

test = [("hi", 1), ("there", 2)]
test = dict(test)
print test["hi"] # prints 1

The filter function can also provide an interesting solution:

result = list(filter(lambda x: x.count(1) > 0, a))

which searches the tuples in the list a for any occurrences of 1. If the search is limited to the first element, the solution can be modified into:

result = list(filter(lambda x: x[0] == 1, a))

Using filter function:

>>> def get_values(iterables, key_to_find):

        return list(filter(lambda x:key_to_find in x, iterables)) 

>>> a = [(1,2),(1,4),(3,5),(5,7)]

>>> get_values(a, 1)

>>> [(1, 2), (1, 4)]

>>> [i for i in a if 1 in i]

[(1, 2), (1, 4)]

Or takewhile, ( addition to this, example of more values is shown ):

>>> a= [(1,2),(1,4),(3,5),(5,7),(0,2)]
>>> import itertools
>>> list(itertools.takewhile(lambda x: x[0]==1,a))
[(1, 2), (1, 4)]
>>> 

if unsorted, like:

>>> a= [(1,2),(3,5),(1,4),(5,7)]
>>> import itertools
>>> list(itertools.takewhile(lambda x: x[0]==1,sorted(a,key=lambda x: x[0]==1)))
[(1, 2), (1, 4)]
>>> 

if you want to search tuple for any number which is present in tuple then you can use

a= [(1,2),(1,4),(3,5),(5,7)]
i=1
result=[]
for j in a:
    if i in j:
        result.append(j)

print(result)

You can also use if i==j[0] or i==j[index] if you want to search a number in particular index