5 回答

TA貢獻(xiàn)1891條經(jīng)驗(yàn) 獲得超3個贊
main.py
a = ['A', 'B', 'C', 'D', 'E']
b = [i + j for i, j in zip(a[:-1:2], a[1::2])]
if len(a) % 2 == 1:
b.append(a[-1])
print(b)
結(jié)果
$ python main.py
['AB', 'CD', 'E']

TA貢獻(xiàn)1898條經(jīng)驗(yàn) 獲得超8個贊
>>> [''.join(a[i:i+2]) for i in range(0, len(a), 2)]
['AB', 'CD', 'E']
或者(因?yàn)槲蚁矚g迭代器)
>>> it = iter(a)
>>> [s + next(it, '') for s in it]
['AB', 'CD', 'E']

TA貢獻(xiàn)1868條經(jīng)驗(yàn) 獲得超4個贊
我認(rèn)為更簡單的方法是迭代數(shù)組并連接字符,如果您有偶數(shù)長度的數(shù)組,它就可以工作,因此您可以添加檢查并在奇數(shù)長度的情況下附加最后一個元素。
array = ['A', 'B', 'C', 'D', 'E']
array2 = [f"{array[i]}{array[i+1]}" for i in range(0, len(array)-1, 2)]
if len(array)%2!=0:
array2.append(array[-1])
print(array2)

TA貢獻(xiàn)1799條經(jīng)驗(yàn) 獲得超9個贊
嘗試這樣。這是非常簡單的答案,但應(yīng)該有效。
my_array = ['A', 'B', 'C', 'D', 'E']
def combine_array(my_array):
mixed_array = []
start_new = True
for item in my_array:
if start_new:
mixed_array.append(item)
start_new = False
else:
mixed_array[-1] = mixed_array[-1] + item
start_new = True
return mixed_array
if __name__ == "__main__":
try:
print(combine_array(my_array))
except Exception as err:
print(err)

TA貢獻(xiàn)1798條經(jīng)驗(yàn) 獲得超3個贊
startArray = ['A', 'B', 'C', 'D', 'E']
currentIndex = 0
finishArray = ['']
for x in startArray:
if len(finishArray[currentIndex]) == 2:
currentIndex += 1
finishArray.insert(currentIndex,x)
else:
finishArray[currentIndex] += x
print(finishArray)
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