我有一个这样的数组：
字段4是1,2,3的平均值，字段5是1,2,3的最小值。
unittest.cfg已按字段4然后按字段5排序。
我希望列举此列表，以创建一种“等级”或“讲台”。
enumerate（） 不起作用，因为如您所见，某些字段绑定在字段4和5上，因此它们的“等级”应该相同。
例如，第一个值应类似于：
[['name0', 24, 19, 25, 22.67, 19],
['name1', 25, 19, 25, 23.0, 19],
['name2', 25, 19, 25, 23.0, 19],
['name3', 24, 22, 23, 23.0, 22],
['name4', 27, 19, 25, 23.67, 19],
['name5', 27, 19, 25, 23.67, 19],
['name6', 28, 19, 26, 24.33, 19],
['name7', 28, 19, 26, 24.33, 19],
['name8', 28, 19, 26, 24.33, 19],
['name9', 26, 22, 27, 25.0, 22],
['name10', 27, 23, 25, 25.0, 23],
['name11', 30, 19, 27, 25.33, 19],
['name12', 24, 31, 28, 27.67, 24],
['name13', 28, 27, 28, 27.67, 27],
['name14', 27, 29, 27, 27.67, 27],
['name15', 29, 26, 29, 28.0, 26],
['name16', 29, 26, 30, 28.33, 26],
['name17', 30, 31, 26, 29.0, 26],
['name18', 33, 27, 30, 30.0, 27],
['name19', 29, 31, 30, 30.0, 29],
['name20', 30, 36, 31, 32.33, 30],
['name21', 36, 30, 32, 32.67, 30],
['name22', 38, 33, 36, 35.67, 33],
['name23', 30, 27, 99, 52.0, 27],
['name24', 99, 27, 32, 52.67, 27],
['name25', 37, 99, 36, 57.33, 36]]
无法找到一种干净的方法来解决这个问题。 感谢您的帮助。
答案 0 :(得分：1)
假定列表已排序，则可以使用恰当命名的groupby和itemgetter按子列表的第4个和第5个元素对子列表进行分组。在enumerate返回的迭代器上使用groupby：
from itertools import groupby
from operator import itemgetter
# data = [['name0', ...
[ [str(i+1)] + l for i, (k, g) in enumerate(groupby(data, key=itemgetter(4, 5))) for l in g ]
输出：
[
['1', 'name0', 24, 19, 25, 22.67, 19],
['2', 'name1', 25, 19, 25, 23.0, 19],
['2', 'name2', 25, 19, 25, 23.0, 19],
['3', 'name3', 24, 22, 23, 23.0, 22],
['4', 'name4', 27, 19, 25, 23.67, 19],
['4', 'name5', 27, 19, 25, 23.67, 19],
['5', 'name6', 28, 19, 26, 24.33, 19],
['5', 'name7', 28, 19, 26, 24.33, 19],
['5', 'name8', 28, 19, 26, 24.33, 19],
['6', 'name9', 26, 22, 27, 25.0, 22],
['7', 'name10', 27, 23, 25, 25.0, 23],
['8', 'name11', 30, 19, 27, 25.33, 19],
['9', 'name12', 24, 31, 28, 27.67, 24],
['10', 'name13', 28, 27, 28, 27.67, 27],
['10', 'name14', 27, 29, 27, 27.67, 27],
['11', 'name15', 29, 26, 29, 28.0, 26],
['12', 'name16', 29, 26, 30, 28.33, 26],
['13', 'name17', 30, 31, 26, 29.0, 26],
['14', 'name18', 33, 27, 30, 30.0, 27],
['15', 'name19', 29, 31, 30, 30.0, 29],
['16', 'name20', 30, 36, 31, 32.33, 30],
['17', 'name21', 36, 30, 32, 32.67, 30],
['18', 'name22', 38, 33, 36, 35.67, 33],
['19', 'name23', 30, 27, 99, 52.0, 27],
['20', 'name24', 99, 27, 32, 52.67, 27],
['21', 'name25', 37, 99, 36, 57.33, 36]
]
答案 1 :(得分：0)
从i = 1开始并遍历它们并分配等级，如果下一行不同，则仅递增i += 1。
答案 2 :(得分：0)
使用Pandas和dense rank：
import pandas as pd
df = pd.DataFrame(data = [['name0', 24, 19, 25, 22.67, 19],
['name1', 25, 19, 25, 23.0, 19],
['name2', 25, 19, 25, 23.0, 19],
['name3', 24, 22, 23, 23.0, 22],
['name4', 27, 19, 25, 23.67, 19],
['name5', 27, 19, 25, 23.67, 19],
['name6', 28, 19, 26, 24.33, 19],
['name7', 28, 19, 26, 24.33, 19],
['name8', 28, 19, 26, 24.33, 19],
['name9', 26, 22, 27, 25.0, 22],
['name10', 27, 23, 25, 25.0, 23],
['name11', 30, 19, 27, 25.33, 19],
['name12', 24, 31, 28, 27.67, 24],
['name13', 28, 27, 28, 27.67, 27],
['name14', 27, 29, 27, 27.67, 27],
['name15', 29, 26, 29, 28.0, 26],
['name16', 29, 26, 30, 28.33, 26],
['name17', 30, 31, 26, 29.0, 26],
['name18', 33, 27, 30, 30.0, 27],
['name19', 29, 31, 30, 30.0, 29],
['name20', 30, 36, 31, 32.33, 30],
['name21', 36, 30, 32, 32.67, 30],
['name22', 38, 33, 36, 35.67, 33],
['name23', 30, 27, 99, 52.0, 27],
['name24', 99, 27, 32, 52.67, 27],
['name25', 37, 99, 36, 57.33, 36]], columns= ['1', '2', '3', '4', '5', '6'])
df["rank"] = df['5'].rank(method = "dense")
df
>
1 2 3 4 5 6 rank
0 name0 24 19 25 22.67 19 1.0
1 name1 25 19 25 23.00 19 2.0
2 name2 25 19 25 23.00 19 2.0
3 name3 24 22 23 23.00 22 2.0
4 name4 27 19 25 23.67 19 3.0
5 name5 27 19 25 23.67 19 3.0
6 name6 28 19 26 24.33 19 4.0
7 name7 28 19 26 24.33 19 4.0
8 name8 28 19 26 24.33 19 4.0
9 name9 26 22 27 25.00 22 5.0
10 name10 27 23 25 25.00 23 5.0
11 name11 30 19 27 25.33 19 6.0
12 name12 24 31 28 27.67 24 7.0
13 name13 28 27 28 27.67 27 7.0
14 name14 27 29 27 27.67 27 7.0
15 name15 29 26 29 28.00 26 8.0
16 name16 29 26 30 28.33 26 9.0
17 name17 30 31 26 29.00 26 10.0
18 name18 33 27 30 30.00 27 11.0
19 name19 29 31 30 30.00 29 11.0
20 name20 30 36 31 32.33 30 12.0
21 name21 36 30 32 32.67 30 13.0
22 name22 38 33 36 35.67 33 14.0
23 name23 30 27 99 52.00 27 15.0
24 name24 99 27 32 52.67 27 16.0
25 name25 37 99 36 57.33 36 17.0
如果要列表列表-
df = df.set_index('rank').reset_index()
df.values.tolist()
答案 3 :(得分：0)
您可以在将None值填充到其中一项后，通过将列表与自身之间的位置进行压缩来对相邻项目进行配对，以便您可以遍历经过压缩的对以比较关键字段，如果它们相同，则可以重复使用以前的排名：
for i, ((*_, prev_mean, prev_min), (*_, mean, _min)) in enumerate(zip([(None, None)] + l, l)):
l[i].insert(0, str(l[i - 1][0] if mean == prev_mean and _min == prev_min else i + 1))
假设列表列表存储为变量l，则l变为：
[['1', 'name0', 24, 19, 25, 22.67, 19],
['2', 'name1', 25, 19, 25, 23.0, 19],
['2', 'name2', 25, 19, 25, 23.0, 19],
['4', 'name3', 24, 22, 23, 23.0, 22],
['5', 'name4', 27, 19, 25, 23.67, 19],
['5', 'name5', 27, 19, 25, 23.67, 19],
['7', 'name6', 28, 19, 26, 24.33, 19],
['7', 'name7', 28, 19, 26, 24.33, 19],
['7', 'name8', 28, 19, 26, 24.33, 19],
['10', 'name9', 26, 22, 27, 25.0, 22],
['11', 'name10', 27, 23, 25, 25.0, 23],
['12', 'name11', 30, 19, 27, 25.33, 19],
['13', 'name12', 24, 31, 28, 27.67, 24],
['14', 'name13', 28, 27, 28, 27.67, 27],
['14', 'name14', 27, 29, 27, 27.67, 27],
['16', 'name15', 29, 26, 29, 28.0, 26],
['17', 'name16', 29, 26, 30, 28.33, 26],
['18', 'name17', 30, 31, 26, 29.0, 26],
['19', 'name18', 33, 27, 30, 30.0, 27],
['20', 'name19', 29, 31, 30, 30.0, 29],
['21', 'name20', 30, 36, 31, 32.33, 30],
['22', 'name21', 36, 30, 32, 32.67, 30],
['23', 'name22', 38, 33, 36, 35.67, 33],
['24', 'name23', 30, 27, 99, 52.0, 27],
['25', 'name24', 99, 27, 32, 52.67, 27],
['26', 'name25', 37, 99, 36, 57.33, 36]]