Question

因此，列表清单是：

[['1', '1', 'tall', 8.425169446611104], ['1', '2', 'tall', 8.917085904771866],     ['1', '3', 'medium', 6.174348482965436], ['2', '1', 'short', 7.8163095871050965], ['2', '2', 'medium', 9.098197168204184], ['2', '3', 'medium', 7.889044875446846], ['3', '1', 'medium', 7.267435895701576], ['3', '2', 'medium', 2.270144244358967], ['3', '3', 'tall', 9.721626339195156], ['4', '1', 'medium', 8.242226639616785], ['4', '2', 'tall', 11.688990983183421], ['4', '3', 'medium', 5.326719989412714]]

所以我会试着解释一下我想在这里做些什么：
如果我们拿出两个列表：

['1'，'1'，'tall'，8.425169446611104]
['2'，'1'，'short'，7.8163095871050965]

如果列表中的第二个值与列表中的另一个列表匹配（在本例中为“1”），您将获取最后一个数字（8.425169446611104）和另一个数字（7.8163095871050965）并将其除以通过数字（2）来获得平均数。

['1'，'X'，'tall'，Y]
['2'，'X'，'短'，Y]
TLDR：如果X'匹配，则将两个Y加在一起并除以Y的数量（平均Y）（因此它可以是两个具有相同X数的列表）

所以我想要的是具有相同X'的 Y的平均值。

Answer 1

这是一种蛮力，我可以更有效地完成它，但是对于你来说启动是一个好地方。

lists = [['1', '1', 'tall', 8.425169446611104], ['1', '2', 'tall', 8.917085904771866],     ['1', '3', 'medium', 6.174348482965436], ['2', '1', 'short', 7.8163095871050965], ['2', '2', 'medium', 9.098197168204184], ['2', '3', 'medium', 7.889044875446846], ['3', '1', 'medium', 7.267435895701576], ['3', '2', 'medium', 2.270144244358967], ['3', '3', 'tall', 9.721626339195156], ['4', '1', 'medium', 8.242226639616785], ['4', '2', 'tall', 11.688990983183421], ['4', '3', 'medium', 5.326719989412714]]

# Creating a dictionary of each "second element" in the sublists and
# compute the 'subtotal' for each of those values:
d = dict()
for i in range(len(lists)):
    d[lists[i][1]] = lists[i][3] + d.get(lists[i][1], 0)

# now do the division:
for k in d.keys():
    count = len([l for l in lists if l[1] == k])
    print 'average for {} is {}'.format(k, d[k]/count)

这应该打印出以下内容：

{'1': 31.751141569034562, '3': 29.111739687020155, '2': 31.974418300518437}

average for 1 is 7.93778539226
average for 3 is 7.27793492176
average for 2 is 7.99360457513

Answer 2

from collections import defaultdict

all_lists = [['1', '1', 'tall', 8.425169446611104],
             ['1', '2', 'tall', 8.917085904771866],
             ['1', '3', 'medium', 6.174348482965436],
             ['2', '1', 'short', 7.8163095871050965],
             ['2', '2', 'medium', 9.098197168204184],
             ['2', '3', 'medium', 7.889044875446846],
             ['3', '1', 'medium', 7.267435895701576],
             ['3', '2', 'medium', 2.270144244358967],
             ['3', '3', 'tall', 9.721626339195156],
             ['4', '1', 'medium', 8.242226639616785],
             ['4', '2', 'tall', 11.688990983183421],
             ['4', '3', 'medium', 5.326719989412714]]

sublists = defaultdict(list)
for sublist in all_lists:
    sublists[sublist[1]].append(sublist[3])

result = {key: sum(sublist) / len(sublist)
          for key, sublist in sublists.items()}

print result
#  {'1': 7.9377853922586405, '3': 7.277934921755039, '2': 7.993604575129609}

Answer 3

试着告诉我这是不是你想要的

list_v = [YOUR LIST HERE]    
previous_numbers = []
for x in list_v:
    if x[1] not in previous_numbers:
        division = 1
        value = x[3]
        for k in list_v:
            if x!=k:
                if x[1] == k[1]:
                    value += k[3]
                    division += 1
        value = value/division
        print x[1]," has medium value: ",value
        previous_numbers.append(x[1])

输出：

1  has medium value:  7.93778539226
2  has medium value:  7.99360457513
3  has medium value:  7.27793492176

不是非常pythonic但它具有新手可读的优势

Python - 比较列表列表中的第二个元素

3 个答案: