Python从两个具有相同长度的列表中获取唯一值和最小值对

Question

我有两个列表：

index = [1,1,1,1,2,2,2,2,3,4,5,5,5,6,7,8,9,10,10,10]
value = [2,3,2,1,2,4,6,8,2,1,5,2,7,2,2,2,1,55,1,11]

相同的长度，但我想要的回报是索引列表中的唯一数字和根据索引的值列表的最小值

结果应该是这样的：

index_result = [1,2,3,4,5,6,7,8,9,10]

value_result = [1,2,2,1,2,2,2,2,1,1]

我试过了：

index = [1,1,1,1,2,2,2,2,3,4,5,5,5,6,7,8,9,10,10,10]
value = [2,3,2,1,2,4,6,8,2,1,5,2,7,2,2,2,1,55,1,11]
index_result = []
value_result = []
#global small_value
j = 0
while j < len(index):
    if j == 0:
        try:
            if index[j] == index[j+1]:
                small_value = min(value[j],value[j+1])
            elif index[j] != index[j+1]:
                index_result.append(index[j])
                value_result.append(value[j])
        except IndexError as e:
            print(e)
            pass
        j = j + 1
        print('small value is for index j ==0')
        print(small_value)
    elif j <len(index) - 1:
        try:
            # if index[j] == index[j-1]:
            #     small_value = min(value[j],value[j-1])
            if index[j] != index[j+1] and index[j] != index[j-1]:
                index_result.append(index[j])
                value_result.append(value[j])
            elif index[j] != index[j+1] and index[j] == index[j-1]:
                index_result.append(index[j])
                value_result.append(small_value)
        except IndexError as e:
            print(e)
            pass
        j = j + 1
        print('small value is for index 0 < j <len(index)')
        print(small_value)
    elif j == len(index) - 1:
        try:
            if index[j] == index[j-1]:
                small_value = min(value[j],value[j-1])
                index_result.append((index[j]))
                value_result.append(small_value)
            elif index[j] != index[j-1]:
                index_result.append(index[j])
                value_result.append(value[j])
        except IndexError as e:
            print(e)
            pass
        j = j + 1
        print('small value is for j = len(index) - 1')
        print(small_value)

print (index_result)
print (value_result)

结果接近预期但仍然错误：

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10] [2, 2, 2, 1, 2, 2, 2, 2, 1, 1]

Answer 1

既然你标记了熊猫

pd.DataFrame([index,value]).T.sort_values([0,1]).drop_duplicates(0)
     0  1
3    1  1
4    2  2
8    3  2
9    4  1
11   5  2
13   6  2
14   7  2
15   8  2
16   9  1
18  10  1

Answer 2

我认为这就是你想要实现的目标

import math

indices = [1,1,1,1,2,2,2,2,3,4,5,5,5,6,7,8,9,10,10,10]
values = [2,3,2,1,2,4,6,8,2,1,5,2,7,2,2,2,1,55,1,11]

result = [math.inf] * max(indices)

for i, v in zip(indices, values):
    result[i-1] = min(result[i-1], v) # Apparently 1 based indexing

print(result)

[1, 2, 2, 1, 2, 2, 2, 2, 1, 1]

Answer 3

如果您的index 像示例中一样单调递增，您可以使用itertools.groupby 尝试这个python 解决方案

from itertools import groupby

d = {k: min([x[1] for x in g]) for k, g in groupby(zip(index, value), 
                                                     lambda x: x[0])}

In [95]: d
Out[95]: {1: 1, 2: 2, 3: 2, 4: 1, 5: 2, 6: 2, 7: 2, 8: 2, 9: 1, 10: 1}

index_result = list(d.keys())

Out[103]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

value_result = list(d.values())

Out[105]: [1, 2, 2, 1, 2, 2, 2, 2, 1, 1]

Answer 4

Numpy 解决方案：

index = np.array([1,1,1,1,2,2,2,2,3,4,5,5,5,6,7,8,9,10,10,10])
value = np.array([2,3,2,1,2,4,6,8,2,1,5,2,7,2,2,2,1,55,1,11])

[value[index == i].min() for i in np.unique(index)]
# [1, 2, 2, 1, 2, 2, 2, 2, 1, 1]