結合在jython中的兩個列表

Question

我想在jython中編寫一個函數，它將結合兩個不同列表中的元素來創建一個單詞。

例如：

的函數有兩個同時列出了名稱爲a和b

如果是["eat", "pray", "love"] 和b是["er", "ing", "d"] ，你在makeWord類型（A，B） 它會返回搭配：

["eater", "praying", "loved"] 

到目前爲止我有：

def makeWords(a,b): 
    a = [] 
    list1 = a 
    b = [] 
    list2 = b 
    new_list = [] 
    for word in list1: 
     new_list.append((list1[i] + list2[i])) 
    return new_list 

但我知道我顯然至少做了一些錯誤的事情。任何幫助將不勝感激！

Answer 1

如何：

map(lambda x: ''.join(x), zip(list1, list2)) 

或者更易讀：

map(''.join, zip(list1, list2)) 

Answer 2

這應該工作以及：

>>> a = ["eat", "pray", "love"] 
>>> b = ["er", "ing", "d"] 
>>> [start + end for start, end in zip(a, b)] 
['eater', 'praying', 'loved'] 

喬爾 - 科內特的timeit代碼：

import timeit 

a = ''' 
a = ["eat", "pray", "love"] 
b = ["er", "ing", "d"] 
[start + end for start, end in zip(a, b)] 
''' 

b = ''' 
a = ["eat", "pray", "love"] 
b = ["er", "ing", "d"] 
map(lambda x: ''.join(x), zip(a, b)) 
''' 

c = ''' 
a = ["eat", "pray", "love"] 
b = ["er", "ing", "d"] 
map(''.join, zip(a, b)) 
''' 

timeit.Timer(a).timeit(number=1000000) 
timeit.Timer(b).timeit(number=1000000) 
timeit.Timer(c).timeit(number=1000000)