Question

我有一个像这样的列表

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

我想要一个这样的csv

 import csv

 with open('Yo456!.csv', 'wb') as out:
     csv_out = csv.writer(out, delimiter=',')
     csv_out.writerow(['id', 'happy:units', 'cloud', 'sad:units', 'rain', 'rating'])
     for row in lis:
         csv_out.writerow(row)

但是我得到了OutPut

Answer 1

编写一个函数来从数据中提取值。

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

将您的功能用于名利和利润

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

Answer 2

您可以使用以下代码展平行并在行上进行迭代，以获得所需格式的CSV。假设您的数据始终采用给定格式。

    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)

Answer 3

你应该看看@jagadeesh mn，因为它看起来更具普遍性。我的答案非常依赖于不可扩展的定位。但是，我发布了一个快速而脏的代码来压扁。

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))

Answer 4

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))])

Answer 5

只要输入列表与您提到的模式完好无损，下面的代码就可以了。我已将这些项目放在列表中，您可以按照您希望的方式将它们转换为csv。尝试一下。

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))

OutPut：

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

从python的元组列表中编写csv

5 个答案: