python,格式化这个列表

a = [(1, 2), (1, 8), (2, 3), (2, 7), (2, 8), (2, 9), (3, 1), (3, 2), (3, 5), (3, 6),  (3, 7), (3, 7), (3, 9)]

x1=None  # here we keep track of the last x we saw
ys=None  # here we keep track of all ys we've seen for this x1

result = [] 

for x,y in a:
    if x != x1:  # this is an x we haven't seen before
        if ys:   # do we have results for the last x?
            result.append( ys ) 
        ys = [ x, '', y ] # initialize the next set of results
        x1 = x
    else:
        ys.append( y ) # add this to the results we are buliding

if ys:
    result.append( ys )  # add the last set of results

print result

步骤1。将列表转换为词典。每个元素都是具有公共键的值列表。(提示:键是每对的第一个值)

步骤2。现在将每个字典格式化为键、空格和值列表。

不完全是你要求的,但也许更容易合作?

>>> from itertools import groupby
>>> L = [(1, 2), (1, 8), (2, 3), (2, 7), (2, 8), (2, 9), (3, 1), (3, 2), (3, 5), (3, 6), (3, 7), (3, 7), (3, 9)]
>>> for key, group in groupby(L, lambda x: x[0]):
...     print key, list(group)
... 
1 [(1, 2), (1, 8)]
2 [(2, 3), (2, 7), (2, 8), (2, 9)]
3 [(3, 1), (3, 2), (3, 5), (3, 6), (3, 7), (3, 7), (3, 9)]

Link to documentation .

编辑:
我想这更像是你想要的:

>>> d = {}
>>> for key, group in groupby(L, lambda x: x[0]):
...     d[key] = [i[1] for i in group]
... 
>>> d
{1: [2, 8], 2: [3, 7, 8, 9], 3: [1, 2, 5, 6, 7, 7, 9]}

如果您绝对希望密钥是一个字符串,您可以这样编码:

d[str(key)] = [i[1] for i in group]

from collections import defaultdict

s = [
    (1,2),(1,8),
    (2,3),(2,7),(2,8),(2,9),
    (3,1),(3,2),(3,5),(3,6),(3,7),(3,7),(3,9)
    ]

D = defaultdict(list)
for a,b in s:
    D[a].append(b)

L = []
for k in sorted(D.keys()):
    e = [str(k),'']
    e.extend(map(str,D[k]))
    L.append(tuple(e))

print L

输出:

[('1', '', '2', '8'), ('2', '', '3', '7', '8', '9'), ('3', '', '1', '2', '5', '6', '7', '7', '9')]

你必须向你的老师解释它是如何工作的;^)