获取字典中具有最小值的键答案

【问题标题】：Get the key with the minimum value in a dictionary of dictionaries获取字典中具有最小值的键
【发布时间】：2021-12-11 17:08:17
【问题描述】：

我必须编写一个函数find_first_read，以便在字典字典中找到特定的读取键，其中包含的值小于 2

期望的输出

find_first_read(some_dict)
'Read4'

我应该在下面的代码中进行哪些更改以获得所需的输出：

some_dict = {'Read1': {'Read3': 0, 'Read2': 1, 'Read5': 1, 'Read4': 0, 'Read6': 29},
            'Read3': {'Read1': 0, 'Read2': 0, 'Read5': 0, 'Read4': 1, 'Read6': 1},
            'Read2': {'Read1': 13, 'Read3': 1, 'Read5': 21, 'Read4': 0, 'Read6': 0},
            'Read5': {'Read1': 39, 'Read3': 0, 'Read2': 1, 'Read4': 0, 'Read6': 14},
            'Read4': {'Read1': 1, 'Read3': 1, 'Read2': 17, 'Read5': 2, 'Read6': 0},
            'Read6': {'Read1': 0, 'Read3': 43, 'Read2': 0, 'Read5': 0, 'Read4': 1}}


def find_first_read(some_dict):
    positions = [] # output variable
    min_value = 2
    for k, v in some_dict.items():
        if v < min_value:
            positions.append(k)
    return positions

【问题讨论】：

为什么想要的输出是Read4？到处都有小于 2 的值。
是因为Read4的所有值都小于2吗？
是的，正确的。 Read4 的所有值都小于 2

标签： python dictionary for-loop

【解决方案1】：

您实际上是在尝试将您的 min_value 与另一个字典进行比较。您必须添加另一个循环才能获取嵌套值。为了使输出更清晰，我将其作为键列表而不是单个键返回以弥补嵌套。

some_dict = {'Read1': {'Read3': 0, 'Read2': 1, 'Read5': 1, 'Read4': 0, 'Read6': 29},
            'Read3': {'Read1': 0, 'Read2': 0, 'Read5': 0, 'Read4': 1, 'Read6': 1},
            'Read2': {'Read1': 13, 'Read3': 1, 'Read5': 21, 'Read4': 0, 'Read6': 0},
            'Read5': {'Read1': 39, 'Read3': 0, 'Read2': 1, 'Read4': 0, 'Read6': 14},
            'Read4': {'Read1': 1, 'Read3': 1, 'Read2': 17, 'Read5': 2, 'Read6': 0},
            'Read6': {'Read1': 0, 'Read3': 43, 'Read2': 0, 'Read5': 0, 'Read4': 1}}


def find_first_read(some_dict):
    positions = [] # output variable
    min_value = 2
    for k, v in some_dict.items():
        for ki, vi in v.items():
            if vi < min_value:
                positions.append([k, ki])
    return positions
    

print(find_first_read(some_dict))

输出：

[['Read1', 'Read3'], ['Read1', 'Read2'], ['Read1', 'Read5'], ['Read1', 'Read4'], 
['Read3', 'Read1'], ['Read3', 'Read2'], ['Read3', 'Read5'], ['Read3', 'Read4'], 
['Read3', 'Read6'], ['Read2', 'Read3'], ['Read2', 'Read4'], ['Read2', 'Read6'], 
['Read5', 'Read3'], ['Read5', 'Read2'], ['Read5', 'Read4'], ['Read4', 'Read1'], 
['Read4', 'Read3'], ['Read4', 'Read6'], ['Read6', 'Read1'], ['Read6', 'Read2'], 
['Read6', 'Read5'], ['Read6', 'Read4']]

【讨论】：

【解决方案2】：

用途：

from collections import defaultdict

def find_first_read(data):

    d = defaultdict(list)
    for key, values in data.items():
        for k, v in values.items():
            d[k].append(v)

    # for example the dictionary has the item "Read4", [0, 1, 0, 0, 1]
    return next(k for k, vs in d.items() if all(v < 2 for v in vs))


some_dict = {'Read1': {'Read3': 0, 'Read2': 1, 'Read5': 1, 'Read4': 0, 'Read6': 29},
             'Read3': {'Read1': 0, 'Read2': 0, 'Read5': 0, 'Read4': 1, 'Read6': 1},
             'Read2': {'Read1': 13, 'Read3': 1, 'Read5': 21, 'Read4': 0, 'Read6': 0},
             'Read5': {'Read1': 39, 'Read3': 0, 'Read2': 1, 'Read4': 0, 'Read6': 14},
             'Read4': {'Read1': 1, 'Read3': 1, 'Read2': 17, 'Read5': 2, 'Read6': 0},
             'Read6': {'Read1': 0, 'Read3': 43, 'Read2': 0, 'Read5': 0, 'Read4': 1}}

res = find_first_read(some_dict)
print(res)

输出

Read4

步骤

在嵌套的 for 循环之后，d 具有以下值：

{'Read1': [0, 13, 39, 1, 0],
 'Read2': [1, 0, 1, 17, 0],
 'Read3': [0, 1, 0, 1, 43],
 'Read4': [0, 1, 0, 0, 1],
 'Read5': [1, 0, 21, 2, 0],
 'Read6': [29, 1, 0, 14, 0]}

最后一行使用next 获取生成器表达式的第一个值：

k for k, vs in d.items() if all(v < 2 for v in vs)

此表达式遍历d 的键、值对并返回所有值都小于2 的键，最后一部分使用函数all 进行检查。

【讨论】：

【解决方案3】：

你可以试试这个：

positions=[]
for k, v in some_dict.items():
  for x, y in v.items():
    if y>=2:
      positions.append(x)
positions = list(set(positions))
output=[]
for k in some_dict:
  if k not in positions:
    output.append(k)
print(output)

输出：

['Read4']

【讨论】：