[英]Get the key with the minimum value in a dictionary of dictionaries
我必须编写一个函数find_first_read
才能在字典中找到包含小于 2 的值的特定读取键
期望的输出
find_first_read(some_dict)
'Read4'
我应该在下面的代码中更改什么以获得所需的输出:
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
您实际上是在尝试将 min_value 与另一个字典进行比较。 您必须添加另一个循环来获取嵌套值。 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']]
用:
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
脚步
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
来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)
此表达式迭代d
的键值对,并返回所有值都小于2
的键,使用函数all
检查最后一部分。
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']
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