从python元组列表编写csv

我有一张这样的清单

lis = [('a', [('happy:units:1', 'cloud:1'), ('sad:units:1', 'rain:2'), 'rating:3']),]

我想要一个像这样的csv

但是我得到了输出

编写一个函数从数据中提取值

def flatten_row(row):
    """
    Transform a row in my weird format into a flat tuple of values.

    >>> ('a', [('happy:units:1', 'cloud:1'), ('sad:units:1', 'rain:2'), 'rating:3'])
    ('a', '1', '1', '1', '2', '3')
    """
    pass  # YOUR CODE HERE

利用你的功用获得名利

for row in lis:
     csv_out.writerow(flatten_row(row))

---

您可以使用下面的代码展平该行，并在该行上进行迭代，以获得所需格式的CSV。假设您的数据始终为给定格式

    lis = [('a', [('happy:units:1', 'cloud:1'), ('sad:units:1', 'rain:2'), 'rating:3']),]
    return_list = list()
    # For the sake of simplicity, Im taking the first element of the given list
    for item in lis[0]:
      # Check if the item is a list
      if type(item) is list:
        for it in item:
          # Iterate over the list item to flatten
          if type(it) is tuple:
            for x in it:
              return_list.append(x[x.rfind(':')+1:])
          else:
            return_list.append(it[it.rfind(':')+1:])
       else:
         # If it is not a list then simply append the value
         return_list.append(item)
    print(return_list)

---

您可能应该看看@jagadeesh mn，因为它看起来更具概括性。我的答案非常依赖于不可扩展的定位。然而，我发布了一个快速而肮脏的代码来扁平化

test=[('bba', [('happy:units:1', 'cloud:1'), ('sad:units:1', 'rain:2'), 'rating:3']),('bcc', [('happy:units:2', 'cloud:2'), ('sad:units:1', 'rain:3'), 'rating:5'])]
col_id=[];
col_rating=[];
col_happy=[];
col_cloud=[];
col_sad=[];
col_rain=[];

for row in test:
    col_id.append(row[0])
    for j,row2 in enumerate(row[1]):
        if(j==2):
            col_rating.append(row2[row2.rfind(':')+1:])
        if(j==0):
            for k,row3 in enumerate(row2):
                if(k==0):
                    col_happy.append(row3[row3.rfind(':')+1:])
                if(k==1):
                    col_cloud.append(row3[row3.rfind(':')+1:])
        if(j==1):
            for k,row3 in enumerate(row2):
                if(k==0):
                    col_sad.append(row3[row3.rfind(':')+1:])
                if(k==1):
                    col_rain.append(row3[row3.rfind(':')+1:])


final_list = list(zip(col_id,col_rating,col_happy,col_cloud,col_sad,col_rain))

---

不易出错的解决方案：

li = [('a', [('happy:units:1', 'cloud:1'), ('sad:units:1', 'rain:2'), 'rating:3']),]

li = list(li[0])

li = [li[0]] + li[1]

def flatten_row(inp_lst):
    ou = []
    for x in inp_lst:
        if isinstance(x, basestring):
            ou.append(x)
        else:
            ou += list(x)
    return ou

def get_values(inp_lst):
    ou = []
    for el in inp_lst:
        pos = el.rfind(":") + 1
        ou.append(el[pos:])
    return ou

with open(output_file, "wb") as f:
    writer = csv.writer(f)
    writer.writerows([['id', 'happy:units', 'cloud', 'sad:units', 'rain', 'rating'], 
    get_values(flatten_row(li))])

---

只要您的输入列表与您提到的模式保持一致，下面的代码就可以工作。我已将项目放入列表中，您可以按照自己的方式将其转换为csv。试试看

lis = [('a', [('happy:units:1', 'cloud:1'), ('sad:units:1', 'rain:2'), 'rating:3']),]
res = list()
for i in xrange(0,len(lis[len(lis)-1])):
        if len(lis[0][i]) > 0:
                for j in xrange(0,len(lis[0][i])):
                        if len(lis[0][i][j]) >= 1 and type(lis[0][i][j]) == tuple:
                                for x in xrange(0,len(lis[0][i][j])):
                                        res.append(lis[0][i][j][x])
                        else:
                                res.append(lis[0][i][j])
csv = list()
csv.append(res[0])
for i in xrange(1,len(res)):
        csv.append(res[i][:-2]+'='+res[i][len(res[i])-1:])
print("Output: {}".format(csv))

输出：

['a', 'happy:units=1', 'cloud=1', 'sad:units=1', 'rain=2', 'rating=3']

---

这是因为源列表只包含两列。第二列依次包含另一个列表/元组。您需要创建一个新列表，从嵌套元组中提取内容，然后将此新列表写入csv。为什么您的数据采用这种格式？这是某个列表的输出吗？是的，这是一个列表的输出。您是否设法获得了展平代码，还是仍然需要答案？我无法展平代码

['a', 'happy:units=1', 'cloud=1', 'sad:units=1', 'rain=2', 'rating=3']