如何将元组的多个级别展平并将元素分开？

Question

这是一个用于计算单词相似度的函数，我是用一个python-excel库xlwings的import编写的。我希望它像这样返回（我所期望的是，每一行/每一行中的项目都应该被一个选项卡分割）。然后我可以轻松地复制/粘贴到Excel文件中，以获得一个总和），例如：

0.9999998807907104 'casual' 1.0 1.0 29.0
0.8386740684509277 'active' 0.3333 1.0 13.0
0.776314377784729 'cardigans'0.1667 1.0 84.0

但它实际上是这样返回的（我讨厌的是，我不能复制到Excel文件以供进一步使用，比如求和数字）：

[[0.9999998807907104, ('casual', (1.0, 1.0, 29.0))],
 [0.8386740684509277, ('active', (0.3333, 1.0, 13.0))],
 [0.776314377784729, ('cardigans', (0.1667, 1.0, 84.0))]]

我怎么知道？谢谢你

def similarity(phrase, N=10):
    phrase_vec = phrase_model[phrase]
    CosDisList = []
    wb = xw.Book('file01.xlsx')
    sht = wb.sheets['sheet1']


    for a_word in phrase_model.keys():

        a_val = phrase_model[a_word]
        cos_dis = cosine_similarity(phrase_vec, a_val)

        for i in range(1, 18):

            if a_word == sht.cells(i, 1).value:
                DataFromExcel = (sht.cells(i, 2).value, sht.cells(i, 3).value, sht.cells(i, 4).value)
                DataCombined = (a_word, DataFromExcel)
                CosDisBind = [float(str(cos_dis.tolist()).strip('[[]]')), DataCombined]

                CosDisList.append(CosDisBind)

                CosDisListSort = sorted(CosDisList, key=operator.itemgetter(0), reverse=True)

                CosDisListTopN = heapq.nlargest(N, CosDisListSort)

    return CosDisListTopN

Answer 1

您可以使用以下函数。资料来源：a blogpost

def flatten(l, ltypes=(list, tuple)):
    ltype = type(l)
    l = list(l)
    i = 0
    while i < len(l):
        while isinstance(l[i], ltypes):
            if not l[i]:
                l.pop(i)
                i -= 1
                break
            else:
                l[i:i + 1] = l[i]
        i += 1
    return ltype(l)

那就用：

abc = [[0.9999998807907104, ('casual', (1.0, 1.0, 29.0))],
       [0.8386740684509277, ('active', (0.3333, 1.0, 13.0))],
       [0.776314377784729, ('cardigans', (0.1667, 1.0, 84.0))]]
flat_list = flatten(abc)
final_array = np.array(flat_list).reshape((np.round(len(flat_list)//5), 5)).tolist()
# [['0.9999998807907104', 'casual', '1.0', '1.0', '29.0'], ['0.8386740684509277', 'active', '0.3333', '1.0', '13.0'], ['0.776314377784729', 'cardigans', '0.1667', '1.0', '84.0']]

现在您可以加入各个列表：

most_final = ["\t".join(x) for x in final_array]
print(most_final[0])

输出

print(most_final[0])
0.9999998807907104  casual  1.0 1.0 29.0