擴展幾個列表優雅

Question

鑑於若干名單：

a = ["a1", "a2", "a3"] 
b = ["b1", "b2", "b3"] 
... 
n = ["n1", "n2", "n3"] 

和新值的列表：

new_vals = ["a4", "b4", "n4"] 

我想獲得：

["a1", "a2", "a3", "a4"] 
["b1", "b2", "b3", "b4"] 
... 
["n1", "n2", "n3", "n4"] 

我可以做到這一點當然，循環和臨時變量。好像來了zip,map和list.extend的組合應該更優雅地做，但是正在逃避我。

Answer 1

事情是這樣的：

a = ["a1", "a2", "a3"] 
b = ["b1", "b2", "b3"] 

# Put the list a, b ... in a big_list. 
big_list = [a, b] 

new_vals = ["a4", "b4", "n4"] 

for i, new_val in enumerate(new_vals): 
    big_list[i].append(new_val) 

Answer 2

假設你有一個列表的列表：

lsts = [ ['a1','a2','a3'], 
     ['b1','b2','b3'], 
     ['c1','c2','c3'] ] 

而且你要追加到每個列表的中端新值的列表lsts：

lst = [ 'a4', 'b4', 'c4' ] 

然後你可以使用列表理解：

new_lsts = [l + [x] for l, x in zip(lsts, lst)] 

Answer 3

的地圖（）取向溶液：

a = ["a1", "a2", "a3"] 
b = ["b1", "b2", "b3"] 
n = ["n1", "n2", "n3"] 
new_vals = ["a4", "b4", "n4"] 

map(lambda lst, x: lst.append(x), (a, b, n), new_vals) 

並且結合上面的列表建議列表：

lsts = [['a1','a2','a3'], 
     ['b1','b2','b3'], 
     ['c1','c2','c3']] 
new_vals = ["a4", "b4", "c4"] 
map(lambda lst, x: lst.append(x), lsts, new_vals) 

，因爲它在就地修改LSTS代替這可能是優選創建一個新的列表清單。