按特定索引合并python列表

Question

我有一个数组，我希望在其中合并一个特定的索引。

["captain:", "robot", "alpha", "beta:", "gama", "delta:", "fighter", "test", "exp"]

这个想法，每个偶数元素都应该包含：而其他元素不应该包含：

我想要的输出：

["captain:", "robot, alpha", "beta:", "gama", "delta:", "fighter, test, exp"]

例如，考虑到阵列是可变的，阵列可以是：

["captain:", "robot", "beta:", "game", "exp", "delta:", "fighter", "test"]

输出应该是相同的，每个偶数带：和奇数索引不带：

有人能帮忙吗？

Answer 1

下面是一个使用groupby:的实现

import itertools
x = ["captain:", "robot", "alpha", "beta:", "gama", "delta:", "fighter", "test", "exp"]
new_list = []
for key, group in itertools.groupby(x, key = lambda string: string.endswith(':')):
if key: # in case you happen to consecutive values that end in a colon
new_list += list(group)
else: # these elements do not end in a colon
new_list.append(', '.join(group))
print(new_list)
# ['captain:', 'robot, alpha', 'beta:', 'gama', 'delta:', 'fighter, test, exp']

这里，分组关键字检查项目是否以":"结束，如果是，则key为True，否则为False。如果key的连续值相同，则将它们放在同一组中。因此，包含冒号的字符串后面的任何数量的元素都将被分组在一起。

Answer 2

itertools.groupby()已经为您介绍了。

lst = ["captain:", "robot", "alpha", "beta:", "gama", "delta:", "fighter", "test", "exp"]
print([', '.join(g[1]) for g in itertools.groupby(lst, lambda s: s.endswith(':'))]) 
# ['captain:', 'robot, alpha', 'beta:', 'gama', 'delta:', 'fighter, test, exp']

Answer 3

我不知道groupby函数。因此，如果有人感兴趣，这里有一种替代方法：

sample = ["captain:", "robot", "alpha", "beta:", "gama", "delta:", "fighter", "test", "exp"]

def find_next_index(sample, start):
for i in range(len(sample)):
if i>start and ":" in sample[i]:
return i, ", ".join(sample[start+1:i]), False

return start, ", ".join(sample[start+1:]), True

result = []
start = 0
while True:
x,y,z = find_next_index(sample, start)
result.append(sample[start])
result.append(y)
start = x
if z:
break

print(result)

输出：

['captain:', 'robot, alpha', 'beta:', 'gama', 'delta:', 'fighter, test, exp']