Question

我有以下列表：

mylist = [('word1', '10', 200), ('word2', '10', 450), ('word1', '136', 300), ('word2', '110', 666)]

我想知道如何获得输出：

[('10', 650), ('136',300), ('110', 666)]

所以将位置[1]和[2]添加到新列表或词典中。但是如果word1和word2具有相同的[1]，则将[1]与[2]的总和相加到新列表或词典中。

我希望我足够清楚，否则请求离开。

Answer 1

请将您的list列表重命名为list_（或其他内容），以便不会覆盖对内置列表的引用。

alrd_seen={}
for l in list_:
    alrd_seen[l[1]] = alrd_seen.get(l[1], 0) + l[2]
print list(alrd_seen.items())

with defaultdict;

from collections import defaultdict as dd
alrd_seen = dd(0)
for l in list_:
    alrd_seen[l[1]] += l[2]
print list(alrd_seen.items())

Answer 2

首先将变量名称从list更改为其他内容，例如lst，因为list混淆了数据类型。尝试类似：

getfunc = lambda x, lst: sum(i[2] for i in lst if i[1] == x)
dct = { y: getfunc(y, lst) for x, y, z in lst}

print (dct)

给出输出：

{'110': 666, '10': 650, '136': 300}

Answer 3

mylist = [('word1', '10', 200), ('word2', '10', 450), ('word1', '136', 300), ('word2', '110', 666)]

# Since you need to refer back to previous values while processing,
# we'll use a dictionary as intermediate storage
nl = {}
# Use unpacking to make code more readable and avoid the use of
# error-prone index-based referencing
for _, key, value in mylist:
    # Use dict.get(key, default_value_if_missing) to retrieve the
    # current value or 0 if it is a newly seen key. Add the current
    # value and store.
    nl[key] = nl.get(key, 0) + value

# dict.items() returns a "set-like" list of key-value pairs.
# Convert it to a regular list to get a list of tuples.
list(nl.items())
# [('136', 300), ('10', 650), ('110', 666)]

基于第一和第二值的列表元素的组合

3 个答案: