Writing a enumerate function using lambda

Question

I received this exercise:

Write a function enumerate that takes a list and returns a list of tuples containing (index,item) for each item in the list

My problem is that I cannot insert the index and value in one or a combination of for loops. This is the code I managed to make:

a = ["a", "b", "c","a","b","c"]
index = 0
for i in a:
    print (index,i)
    index+=1

This is roughly the code I want to produce (must be on one line):

my_enumerate = lambda x :[(t) for t in x )]
print list(my_enumerate(range(4)))

How can I put it all one lambda line to get (value, index) back? The output should look like:

[(0, "a"), (1, "b"), (2, "c")]

Am I missing something or are you trying to reinvent the enumerate() function? — Martijn Pieters
– Martijn Pieters, Commented May 18, 2015 at 12:49
Or, if it has to be a lambda: my_enumerate = lambda L: list(ind,item for ind,item in enumerate(L)) — Rick
– Rick, Commented May 18, 2015 at 13:33

Padraic Cunningham · Accepted Answer · 2015-05-18 14:07:15Z

5

If you can actually index just add the value by indexing:

my_enumerate = lambda x :[(t, x[t]) for t in range(len(x))]
print list(my_enumerate(a))
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'a'), (4, 'b'), (5, 'c')]

If not use zip and put range in the lambda:

my_enumerate = lambda x: zip(range(len(x), x))
print list(my_enumerate(a))

edited May 18, 2015 at 14:07

answered May 18, 2015 at 12:48

Padraic Cunningham

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3 Comments

jonrsharpe Over a year ago

Your two examples seem quite different - the first expects the length of the list (and the list itself should also be in scope) and the second the list itself. Why not lambda a: [(t, a[t]) for t in range(len(a))]?

Padraic Cunningham Over a year ago

@jonrsharpe, pasted the wrong code. Actually irrelevant as the OP cannot index

jonrsharpe Over a year ago

I read that as meaning that they were unable to do so, rather than prohibited from it; you may well be right.

IKavanagh · Accepted Answer · 2015-11-13 08:46:06Z

1

[(i,a[i])for i in range(len(a))]

edited Nov 13, 2015 at 8:46

IKavanagh

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answered Nov 13, 2015 at 8:24

Gareema

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Comments

Simone Bronzini · Accepted Answer · 2015-05-18 12:52:11Z

0

my_enumerate = lambda x: [(i, x[i]) for i in xrange(len(x))]
a = ["a", "b", "c", "a", "b", "c"]
print my_enumerate(a)

outputs:

[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'a'), (4, 'b'), (5, 'c')]

EDIT: use range instead of xrange and print(...) instead of print if you are using python3

answered May 18, 2015 at 12:52

Simone Bronzini

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1 Comment

Padraic Cunningham Over a year ago

@H_meir, I thought you said you cannot index?

Shashank · Accepted Answer · 2015-05-18 13:20:32Z

0

Python's zip and range functions solve this problem pretty handily.

my_enumerate = lambda seq: zip(range(len(seq)), seq)

In Python 2.x, you should use itertools.izip, and xrange instead.

answered May 18, 2015 at 13:20

Shashank

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Comments

RoadieRich · Accepted Answer · 2015-05-18 13:52:05Z

You could also do it recursively:

>>> myenumerate = lambda l, n=0: [] if not l else (lambda ll = list(l): [(n, ll.pop(0))] + myenumerate(ll, n+1)()

list.pop(n) returns the nth value form the list, and returns it.

The only problem is that you must pass in a list:

>>> myenumerate([1,2,3,4,5,6,7,8])
[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)]
>>> myenumerate("astring")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 1, in <lambda>
AttributeError: 'str' object has no attribute 'pop'
>>> myenumerate(list("astring"))
[(0, 'a'), (1, 's'), (2, 't'), (3, 'r'), (4, 'i'), (5, 'n'), (6, 'g')]

However, if you just blindly added calls to list, you would't be able to replicate the functionality required without using a slice.

A neat trick for bypassing this requirement is to use another lambda:

>>> myenumerate = lambda l, n=0: [] if not l else (lambda ll: [(n, ll.pop(0))] + myenumerate(ll, n+1))(list(l))

>>> myenumerate("astring")
[(0, 'a'), (1, 's'), (2, 't'), (3, 'r'), (4, 'i'), (5, 'n'), (6, 'g')]

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