How to get each item in a list of lists?

Question

I am making a heart game for my assignment, but I don’t know how to get each item in the list list:

>>>Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C],["JH"]],[["7D"]]] 

and what comes to my mind:

 for values in cards: for value in values: 

But I think that I have an element that has 2 lists. How to calculate the one that has 3 and 1 list in cards?

Answer 1

Like this:

 >>> Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]] >>> from compiler.ast import flatten >>> flatten(Cards) ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D'] 

As noted, nacholibre compiler package is deprecated. This is the source of flatten :

 def flatten(seq): l = [] for elt in seq: t = type(elt) if t is tuple or t is list: for elt2 in flatten(elt): l.append(elt2) else: l.append(elt) return l 

Answer 2

Slightly hides oneliner:

 >>> [a for c in Cards for b in c for a in b] ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7', 'D'] 

You might want to give a, b, and c more descriptive names.

Answer 3

If your cards are embedded in an inconvenient way:

 >>> Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]] >>> def getCards(cardList,myCards=[]): #change this to myCards, and pass in a list to mutate this is just for demo if isinstance(cardList,list): for subList in cardList: getCards(subList) else: myCards.append(cardList) return myCards >>> getCards(Cards) ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D'] 

Will recursively go through the list and find all the elements. These are some points that I have performed comparing the performance of the selected flattern method with mine:

 >>> print(timeit.timeit(r'getCards([[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]],[])',setup="from clas import getCards")) 5.24880099297 >>> timeit.timeit(r'flatten([[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]])',setup="from compiler.ast import flatten") 7.010887145996094 

Answer 4

Your list is an incomplete nested list, so you can first make it rectangular using the procedure described here , and then smooth out the resulting numpy.ndarray .

The "ifs" below is also not needed if the last element ['7D'] was [['7D']] (then other answers will also work).

 import numpy as np collector = np.zeros((3,3,3),dtype='|S20') for (i,j,k), v in np.ndenumerate( collector ): try: if not isinstance(cards[i], str): if not isinstance(cards[i][j], str): collector[i,j,k] = cards[i][j][k] else: collector[i,j,0] = cards[i][j] else: collector[i,0,0] = cards[i] except IndexError: collector[i,j,k] = '' print collector[collector<>''].flatten() 

Answer 5

Using generators, you can write a much more readable implementation of flatten :

 def flatten(l): if isinstance(l, list): for e1 in l: for e2 in flatten(e1): yield e2 else: yield l 

Or, if you are using Python 3.3, which has added yield from syntax:

 def flatten(l): if isinstance(l, list): for e in l: yield from flatten(e) else: yield l 

Result:

 >>> list(flatten([[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],[["7D"]]])) ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D'] 

Answer 6

Use 2 nested itertools.chain to flatten the list:

 In [32]: Cards Out[32]: [[['QS', '5H', 'AS'], ['2H', '8H'], ['7C']], [['9H', '5C'], ['JH']], ['7D']] In [33]: from itertools import chain In [34]: [k for k in chain.from_iterable([i for i in chain.from_iterable(Cards)])] Out[34]: ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7', 'D'] 

Answer 7

This solution is very reliable for any type of nested lists or tuples (to add other duplicate types just add another or isinstance(...) in the code below.

It simply calls a function recursively, which expands itself:

 def unfold(lst): output = [] def _unfold(i): if isinstance(i, list) or isinstance(i, tuple): [_unfold(j) for j in i] else: output.append(i) _unfold(lst) return output print unfold(cards) #['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D'] 

Answer 8

Using the Smooth list from Rosetta code, which you could do:

 >>> def flatten(lst): return sum( ([x] if not isinstance(x, list) else flatten(x) for x in lst), [] ) >>> Cards = [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],["7D"]] >>> flatten(Cards) ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D'] >>> 

The solution only aligns the nested lists — not tuples or rows.

Answer 9

 from itertools import chain, imap l= [[["QS","5H","AS"],["2H","8H"],["7C"]],[["9H","5C"],["JH"]],[["7D"]]] k = list(chain.from_iterable(imap(list, l))) m = list(chain.from_iterable(imap(list, k))) print m 

: ['QS', '5H', 'AS', '2H', '8H', '7C', '9H', '5C', 'JH', '7D']

Itertools is awesome!