例如:
["apple", "banana", "apple", "mango"]
字符串“join”方法在这里可能对你有用,但它不会给你你想要的答案。用“nothing”连接列表的元素:
In [1]: "".join(["I","a","m","h","a","p","p","y"])
Out[1]: 'Iamhappy'
如果您的列表包含空格,您可以这样做:
In [2]: "".join(["I"," ","a","m"," ","h","a","p","p","y"])
Out[2]: 'I am happy'
后跟空格上的字符串“split”:
In [3]: 'I am happy'.split(" ")
Out[3]: ['I', 'am', 'happy']
但是解析字典单词的原始结果(Out[1])是另一回事。
您对如何组合列表中的元素有定义吗?
所以,假设你有你的例子
x = ["I","a","m","h","a","p","p","y"]
comb = [1, 2, 5]
然后
def combine(l, comb):
x = []
if sum(comb) == len(l):
for n, i in enumerate(comb):
d = sum(comb[:n])
x.append(''.join(l[d : d + i]))
return x
return l
并将combine(x, comb)
返回['I', 'am', 'happy']
。