Question

这是示例数据。输入目录中的所有内容都是动态的。唯一的问题是数据字典将为input_dict中的每个键值固定7个不同的值。而且它可能只有1个或0个值。

input_dict = { 'all_val' : ['a', 'b', 'c' ],
               '2nd_group' : ['a', 'b'] ,
               '3rd_grp' : ['a' , 'c']}
data = {  
'a' :      [1,0,1,0,0,0,1],
'b' :      [0,0,1,1,0,1,0],
'c' :      [0,1,1,0,0,0,1]    }

required_output = {'2nd_group': 5, '3rd_grp': 4, 'all_val': 6}

逻辑：对于all_val，取a，b和c转到数据字典。如果a [0]，b [0]，c [0]中的任何一个为1，则应考虑1。a [1]，b [1]，c [1]的方式相同，最后对所有1计数

我的解决方案：

temp_dict = {}
output_dict = {}

for a in input_dict.keys():
    temp_dict[a] = [0]*7

for key, value in input_dict.items():
    for v in value:
        for j , d in enumerate(data[v]):
            temp_dict[key][j] = max( temp_dict[key][j] , d  )

for k,v in temp_dict.items():
    total = 0
    for t in temp_dict[k]:
        total = total + t
    output_dict[k] = total

print output_dict

是否有任何方法可以改善性能或解决此问题的其他方法。

Answer 1

from collections import defaultdict

input_dict = { 'all_val' : ['a', 'b', 'c' ],
               '2nd_group' : ['a', 'b'] ,
               '3rd_grp' : ['a' , 'c']}
data = {  
'a' :      [1,0,1,0,0,0,1],
'b' :      [0,0,1,1,0,1,0],
'c' :      [0,1,1,0,0,0,1]    }

# {'2nd_group': 5, '3rd_grp': 4, 'all_val': 6}

temp_dict = defaultdict(list)

SIZE_OF_LIST = 7

data_keys = data.keys()

# we're basically constructiing the temp_dict on the fly by iterating throug the X and Y axis of the matrix
for i in range(SIZE_OF_LIST):  # i is in X axis of the matrix and represents the columns in this case
    for group, group_items in input_dict.items():  # for each column we iterate over the Y axis (a, b, c)

        # we then need to obtain all the values on a column (the actual 0's and 1's) and create a
        # list from it. In this list we take only does rows that are of interest for us
        # For example, for 2nd_group (a, b), considering that we are on column 0 the resulting list
        # will be generated by getting the values for 'a' and 'b', hence we will have [1, 0]
        data_values = [data[data_key][i] for data_key in group_items]  # thanks to list comprehensions

        # we then need to evaluate the previously created list with the any
        # any(data_vaues) is actually any([1, 0]) (from the previous example)
        # which yelds 1, because there is at least one value with the value 1
        # the resulting value is added at the right position in the temp_dict
        temp_dict[group].append(1 if any(data_values) else 0)

output_dict = {}
for group, elements in temp_dict.items():
    # we just iterate over the temp_dict one more time and create the 
    # sums for all our groups (all_val, 2nd_group, 3rd_group)
    # and add up all the 1's in the list. 
    # For example if we're on '2nd_group' then it's basically a sum(temp_dict['2nd_group'])
    # which yields your desired result
    output_dict[group] = sum(elements)

print output_dict

Answer 2

在我的评论之后，有几个部分可以简化：

代替

for k,v in temp_dict.items():
    total = 0
    for t in temp_dict[k]:
        total = total + t
    output_dict[k] = total

您可以写：

output_dict = {k: sum(v) for k,v in temp_dict.items()}

代替

for key, value in input_dict.items():
    for v in value:
        for j , d in enumerate(data[v]):
            temp_dict[key][j] = max( temp_dict[key][j] , d  )

您可以写：

for key, value in input_dict.items():
    temp_dict[key] = [max(data[v][index] for v in value) for index in range(7)]

然后，您可以考虑将所有内容组合在一起并进入：

output_dict = {k: sum(max(data[key][index] for key in keys) for index in range(7)) for k, keys in input_dict.items()}

Answer 3

您可以进行一些调整并简化逻辑。例如，您不需要在第一遍中单独创建密钥。您可以跳过第二遍中的温度指令。整体逻辑可以简化。

input_dict = { 'all_val' : ['a', 'b', 'c' ],
               '2nd_group' : ['a', 'b'] ,
               '3rd_grp' : ['a' , 'c']}
data = {  
'a' :      [1,0,1,0,0,0,1],
'b' :      [0,0,1,1,0,1,0],
'c' :      [0,1,1,0,0,0,1]    }

#required_output = {'2nd_group': 5, '3rd_grp': 4, 'all_val': 6}
res = {} 
for key,value in input_dict.items():
    output = 0
    #create a zip from the lists in data so you can check for 1s at every index
    for i in zip(*[data[v] for v in value]): 
        if any(i): #checking if any of them have a 1. 
            output += 1
    res[key] = output

timeit结果：

新代码：6.36 µs ± 115 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

有问题的代码（基准测试）：19.8 µs ± 339 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Answer 4

您可以使用OR逻辑操作执行以下操作：

import numpy as np
output = {}
for key in input_dict:
    r = []
    for data_key in data:
        if data_key in input_dict[key]:
            if len(r) == 0:
                r = np.asarray(data[data_key])
            else:
                r = r | np.asarray(data[data_key])
    output[key] = list(r).count(1)
print output

Answer 5

我的方法使用7个元素并行计算列表中的所有项目，并且不需要将单独的已安装项目作为numpy。在Python 3中，其内容为：

import operator
import functools

input_dict = { 'all_val' : ['a', 'b', 'c' ],
               '2nd_group' : ['a', 'b'] ,
               '3rd_grp' : ['a' , 'c']}
data = {
    'a' : 0b1010001,
    'b' : 0b0011010,
    'c' : 0b0110001}

def num_bits(n):
    result = 0
    while n > 0:
        result += n & 1
        n >>= 1
    return result

if __name__ == '__main__':
    result = {}
    for inkey, labels in input_dict.items():
        result[inkey] = num_bits(functools.reduce(operator.__or__, (data[l] for l in labels)))
    print(result)

完全冒险的人甚至可以用字典理解代替主要部分：

print({inkey: num_bits(functools.reduce(operator.__or__, (data[l] for l in labels))) for inkey, labels in input_dict.items()})

解决python字典问题的更好方法

5 个答案: