Question

我有两个不同的嵌套列表x和y，它们包含来自巨大语料库的单词的unigrams值。演示列表如下所示：

x = [['ali', '225'], ['raheem', '4514'], ['mohammed', '19652']]
y = [['ali', '45514'], ['mohammed', '441'], ['salwa', '41']]

如上所示，可以在x和y嵌套列表中找到单词，但它们的值不同。使用Python3.3，我尝试比较两个列表，并将y中单词的unigram值添加到x中的相应单词。但是，我收到一条消息错误：

Traceback (most recent call last):
File "<pyshell#120>", line 1, in <module>
add_unigrams(x, y)
File "C:/Python33/add_unigrams.py", line 9, in add_unigrams
x[i][0] = x[i][0], x[i][1] = int(x[i][1]) + int(y[i][1])
TypeError: 'int' object is not iterable

感谢您在使用该代码时提供的任何帮助。这是我的代码：

def add_unigrams(x, y):
'''(lst, lst) -> lst
Compare the items of y to items of x, add the unigrams of similar words
in y to the value of the corresponding words in x. If an item in y is not found 
in x, then x.append(item).

>>> add_unigrams(x, y)
[['ali', '45739'], ['raheem', '4514'], ['mohammed', '20093'], ['salwa', '41']]
'''

i = 0
j = 0

for i in range(len(x)):
    for j in range(len(y)):
        if x[i][0] == y[j][0] and len(x[i]) == len(y[j]):
            x[i][0] = x[i][0]
            x[i][1] = int(x[i][1]) + int(y[i][1])
            i = i + 1
            j = j + 1
    for item in x:
        for item in y:
            if (not item in x):
                x.append(item)


return x

我想实现两个字典来完成同样的任务，但我不知道该怎么做。示例：

d1 = { 'ali': 225, 'raheem' : 4514, 'mohammed' : 19652}
d2 = { 'ali': 45514, 'mohammed' : 441, 'salwa' : 41}

非常感谢任何帮助！穆罕默德

Answer 1

IIUC，而不是使用列表，我使用collections.Counter对象。例如：

>>> x = [['ali', '225'], ['raheem', '4514'], ['mohammed', '19652']]
>>> y = [['ali', '45514'], ['mohammed', '441'], ['salwa', '41']]
>>> x = [[k, int(v)] for k,v in x]
>>> y = [[k, int(v)] for k,v in y]
>>> Counter(dict(x))
Counter({'mohammed': 19652, 'raheem': 4514, 'ali': 225})
>>> Counter(dict(x)) + Counter(dict(y))
Counter({'ali': 45739, 'mohammed': 20093, 'raheem': 4514, 'salwa': 41})

在python.3中从两个嵌套列表中添加unigram值

1 个答案: