如何用另一个列表中的多个元素替换列表中的一个元素？

Question

有两个列表：

list_a = [['apple', 'banana', 'strawberry'], ['meat'], ['milk'], ['meat']]
list_b = [['chicken'], ['pork'], ['beef']]

如何将list_b插入到list_a中，而不是像这样插入“meat”：

list_c = [['apple', 'banana', 'strawberry'], ['chicken'], ['pork'],
['beef'], ['milk'], ['chicken'], ['pork'], ['beef']]

- tmsblgh

Answer 1

使用for循环和.extend()列表方法：

for food in list_a:
    if food == ['meat']:
        list_c.extend(list_b)
    else:
        list_c.append(food)

print (list_c)

而 list_c 将会输出：

[['apple', 'banana', 'strawberry'], ['chicken'], ['pork'], ['beef'], ['milk'], ['chicken'], ['pork'], ['beef']]

你可以使用以下方法将list_a中所有的['meat']替换为list_b中的所有元素。这样就可以得到你想要的输出结果。

- Darren Haynes

Answer 2

切片赋值。

>>> list_c = list_a[:]
>>> list_c[-1:] = list_b[:]
>>> list_c
[['apple', 'banana', 'strawberry'], ['chicken'], ['pork'], ['beef']]

- Ignacio Vazquez-Abrams

Answer 3

>>> idx = list_a.index(['meat'])
>>> list_c = list_a[:idx] + list_b + list_a[idx + 1:]
[['apple', 'banana', 'strawberry'], ['chicken'], ['pork'], ['beef'], ['milk']]

- AKS

Answer 4

def replace(index, L1, L2):
    return L1[0:index] + L2 + L1[index+1:]

只需使用 slice。

- BlackMamba

Answer 5

针对您最后的更改..：

list_a = [['apple', 'banana', 'strawberry'], ['meat'], ['milk'], ['meat']]
list_b = [['chicken'], ['pork'], ['beef']] 
list_c = []
for x in list_a:
    if x == ["meat"]:
        for y in list_b:
            list_c.append(y)
    else:
        list_c.append(x)

list_c
[['apple', 'banana', 'strawberry'], ['chicken'], ['pork'], ['beef'], ['milk'], ['chicken'], ['pork'], ['beef']]

- Dovi

Answer 6

list_a = [['apple', 'banana', 'strawberry'], ['meat'], ['milk']]
list_b = [['chicken'], ['pork'], ['beef']]

deletitem = ['meat']

def newlist(deletetemlist,list_a,list_b):
    deletitemidx = list_a.index(deletetemlist)
    print(list_a[:deletitemidx] + list_b + list_a[deletitemidx + 1:])

newlist(deletitem,list_a,list_b)

- backtrack

---

@AKS，谢谢并修改了它。 - backtrack