Question

我有一个类型为-的字典

{
 'a': [0, 0, 0, 1, 1, 1],
 'diff': [-3.654748632042965, -3.97342251742816, -4.010696955187965, -6.3903113605617685, -6.310284875983931, -4.0363808384515005],
 'b': [-1, 0, 1, -1, 0, 1]
}

我想选择与最小'diff'相对应的a和b值，在这种情况下，它们应为与-6.3903相对应的[1，-1]。我是python新手，无法弄清楚！

Answer 1

您可以使用numpy.argmin()

import numpy as np

d = {'a': [0, 0, 0, 1, 1, 1], 'diff': [-3.654748632042965, -3.97342251742816, -4.010696955187965, -6.3903113605617685, -6.310284875983931, -4.0363808384515005], 'b': [-1, 0, 1, -1, 0, 1]}

ind = np.argmin(d["diff"])

a = d["a"][ind]
b = d["b"][ind]

print([a, b])

OUT：

[1, -1]

Answer 2

使用min查找列表的最小值，然后使用index查找最小值的索引，然后从a和{{1}中获取相应索引的元素}

输出

d={'a': [0, 0, 0, 1, 1, 1], 'diff': [-3.654748632042965, -3.97342251742816, -4.010696955187965, -6.3903113605617685, -6.310284875983931, -4.0363808384515005], 'b': [-1, 0, 1, -1, 0, 1]}
a=d['a']
b=d['b']
diff=d['diff']
index=diff.index(min(diff))
print([a[index],b[index]])

Answer 3

这很简单，如果您知道如何处理字典。这是我在python中的示例代码：

def diff(dict):
    min = 0
    position = 0
    for index in range(len(dict["diff"])):
        checkValue = dict["diff"][index]
        if (checkValue < min):
            min = checkValue
            position = index
    print([dict["a"][position],dict["b"][position]])

测试：

diff({'a': [0, 0, 0, 1, 1, 1], 'diff': [-3.654748632042965, -3.97342251742816, -4.010696955187965, -6.3903113605617685, -6.310284875983931, -4.0363808384515005], 'b': [-1, 0, 1, -1, 0, 1]})

输出：

[1，-1]

Answer 4

您可以将min用于zip的拆包：

from operator import itemgetter

a, diff_min, b = min(zip(*itemgetter('a', 'diff', 'b')(d)), key=itemgetter(1))

print(a, diff_min, b)

1 -6.3903113605617685 -1

选择与键对应的字典值

4 个答案:

测试：