Question

此代码有效，但我想知道是否还有更Python的方式编写它。

word_frequency是列表的字典，例如：

word_frequency = {'dogs': [1234, 4321], 'are': [9999, 0000], 'fun': [4389, 3234]}

vocab_frequency = [0, 0] # stores the total times all the words used in each class
for word in word_frequency: # that is not the most elegant solution, but it works!
    vocab_frequency[0] += word_frequency[word][0] #negative class
    vocab_frequency[1] += word_frequency[word][1] #positive class

是否有更优雅的方式编写此循环？

Answer 1

我不确定这是否更适合Pythonic：

>>> word_frequency = {'dogs': [1234, 4321], 'are': [9999, 0000], 'fun': [4389, 3234]}
>>> vocab_frequency = [sum(x[0] for x in word_frequency.values()),
                       sum(x[1] for x in word_frequency.values())]
>>> print(vocab_frequency)
[15622, 7555]

带有reduce的替代解决方案：

>>> reduce(lambda x, y: [x[0] + y[0], x[1] + y[1]], word_frequency.values())
[15622, 7555]

Answer 2

您可以为此使用numpy：

import numpy as np

word_frequency = {'dogs': [1234, 4321], 'are': [9999, 0000], 'fun': [4389, 3234]}
vocab_frequency = np.sum(list(word_frequency.values()), axis=0)

Answer 3

list(map(sum, zip(*word_frequency.values())))

Answer 4

也许不是解决这个问题的最短方法，但希望是最容易理解的...

word_frequency = {'dogs': [1234, 4321], 'are': [9999, 0000], 'fun': [4389, 3234]}

negative = (v[0] for v in word_frequency.values())
positive = (v[1] for v in word_frequency.values())
vocab_frequency = sum(negative), sum(positive)

print (vocab_frequency)  # (15622, 7555)

尽管经验丰富的Pythonista使用者可能宁愿使用zip来解压缩值：

negative, positive = zip(*word_frequency.values())
vocab_frequency = sum(negative), sum(positive)

Answer 5

另一种方法是：

vocab_frequency[0], vocab_frequency[1] = list(sum([word_frequency[elem][i] for elem in word_frequency]) for i in range(2))

print(vocab_frequency[0])
print(vocab_frequency[1])

输出：

15622
7555

还有，做这件事的另一种方法，有点牵强：

*vocab_frequency, = list(map(sum,zip(*word_frequency.values())))

print(vocab_frequency)

输出：

[15622, 7555]

Answer 6

for frequencies in word_frequency.values():
    vocab_frequency = [sum(x) for x in zip(vocab_frequency, frequencies)]

Answer 7

您可以将该词典转换为pandas DataFrame，它将更容易处理。

import pandas as pd
word_frequency = {'dogs': [1234, 4321], 'are': [9999, 0000], 'fun': [4389, 3234]}

#Syntax to create DataFrame
df = pd.DataFrame(word_frequency)

#Result
   dogs   are   fun
0  1234  9999  4389
1  4321     0  3234

现在只需取每一行的总和，然后转换回列表或保留为数据框对象。

#Take sum of each row and convert to list
df = df.sum(axis=1)
df = df.values.tolist()
print(df)

#Output
[15622, 7555]

Answer 8

尝试此单行解决方案：

[sum([word_frequency[i][0] for i in word_frequency]),sum([word_frequency[i][1] for i in word_frequency])]

Answer 9

for n, p in your_dict.vales():
    res[0] += n
    res[1] += p

这将足够快速而优雅。通过电话发送。抱歉，格式。

在列表字典上执行循环的Python方式

9 个答案: