Question

如何从数据框中获取以下列表列表输出：

输入数据框：

     0          1
0   if         IN
1   trade      NN   
2   figures    NNS
3   for        IN
4   September  NNP
5   ,          ,
6   due        JJ   
7   for        IN
8   release    NN
9   tomorrow   NN
10  ,          ,

输出列表：

[[('if', 'IN'),
 ('trade', 'NN'),
 ('figures', 'NNS'),
 ('for', 'IN'),
 ('September', 'NNP'),
 (',', ',')],
 [('due', 'JJ'),
 ('for', 'IN'),
 ('release', 'NN'),
 ('tommorow', 'NN'),
 (',', ',')]]

只要有定界符,

，该列表就会拆分

Answer 1

修改后的答案：基于关于一旦存在',', ','分隔符时进行拆分的评论：

outputList = list([[]])
ind = 0
for value in df.values:
    value = tuple(value)
    outputList[ind].append(value)
    if (value == (',', ',')):
        outputList.append(list([]))
        ind += 1

#remove last empty inner list which is empty    
outputList.pop()

输出：

[[('if', 'IN'), ('trade', 'NN'), ('figures', 'NNS'), ('for', 'IN'), ('September', 'NNP'), (',', ',')], [('due', 'JJ'), ('for', 'IN'), ('release', 'NN'), ('tomorrow', 'NN'), (',', ',')]]

第一个答案：

为了将数据框转换为元组数组，您可以：

outputList = [[tuple(value) for value in df.values]]
outputList

输出：

[[('if', 'IN'), ('trade', 'NN'), ('figures', 'NNS'), ('for', 'IN'), ('September', 'NNP'), (',', ','), ('due', 'JJ'), ('for', 'IN'), ('release', 'NN'), ('tomorrow', 'NN'), (',', ',')]]

Answer 2

使用：

local function evenly(number, items)
    local space = math.floor((#items-1) / (number-1))
    for i=1,#items,space do
        print(items[i])
    end
end

evenly(5, {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20})

将df转换为列表列表

2 个答案: