如何从 python 中的列表列表创建字典？

Question

Given a list of lists, how can I create a dictionary where the keys are all the items in the list (one copy), and the values are the number of times they are the first item in the list?

Given:

[['banana', 'oranges', 'grapes'],['banana', 'grapes'],['grapes', 'oranges', 'banana']]

Expected:

{'banana': 2, 'grapes': 1, 'oranges': 0}

Answer 1

First get the list of first elements:

filtered_list = [x[0] for x in initial_list]

And here are your unique elements:

unique_elements = set([y for x in initial_list for y in x])

Now initialize a dictionary with keys from unique elements and values of zero:

counts = {e: 0 for e in unique_elements}

Finally update the values of the dictionary with the frequencies of each element in the filtered list. Since you asked a solution without counter , here is a simple loop to achieve that:

for i in filtered_list:
    counts[i] = counts.get(i, 0) + 1

print(counts)
# {'banana': 2, 'grapes': 1, 'oranges': 0}

Answer 2

You can use sets to get unique names present in the sub lists:

initial_list = [['banana', 'oranges', 'grapes'],['banana', 'grapes'],['grapes', 'oranges', 'banana']]

unique = set()

for l in initial_list:
    unique = unique.union(set(l))

Then counting in how many list each item is present (assuming each item is either present or not, not duplicated):

from collections import defaultdict

result = defaultdict(lambda: 0)
for element in unique:
    for l in initial_list:
        result[element] += (element == l[0])

Defaultict is used to get an initial value of 0 And you should have your result in result

The fact that bool are a subclass of int is used to evaluate element == l[0] either to 1 or 0

Without collections you would need to edit the last line to be:

try:
    result[element] += (element == l[0])
except KeyError:
    result[element] = 1

Answer 3

Create list of lists:

ll = [['banana', 'oranges', 'grapes'], ['banana', 'grapes'], ['grapes', 'oranges', 'banana']]

Get unique keys:

from itertools import chain

d = dict.fromkeys(chain(*ll), 0)

Count first elements of lists:

from collections import Counter
from operator import itemgetter

c = Counter(map(itemgetter(0), ll))

Update and show result:

d.update(dict(c))
print(d)

Prints:

{'banana': 2, 'oranges': 0, 'grapes': 1}

Answer 4

All the previous responses failed for me. This is a very fast implementation that works:

# l is a list of lists
import itertools
ext_l = list(itertools.chain.from_iterable(l))
l_dic = {element:0 for element in set(ext_l)}
for i in ext_l: l_dic[i] += 1
print(l)

Answer 5

Simple implementation

l = [['banana', 'oranges', 'grapes'],['banana', 'grapes'],['grapes', 'oranges', 'banana']]
unique_items = set([i for sl in l for i in sl])
d = dict()
for item in unique_items:
    d[item] = 0

for sublist in l:
    d[sublist[0]] += 1

print(d)
# output
# {'grapes': 1, 'oranges': 0, 'banana': 2}

To maintain order

d = dict()
for sl in l:
    d[sl[0]] = d.get(sl[0],0) + 1
print(d)
# {'banana': 2, 'grapes': 1}

unique_items = set([i for sl in l for i in sl])
for item in unique_items:
    if item not in d:
        d[item] = 0
print(d)
# {'banana': 2, 'grapes': 1, 'oranges': 0}