Question

我如何遍历每个嵌套列表的内容？我显然可以在关键字中使用I，但这只是按顺序打印整个列表。我需要做的是循环遍历每个列表的内容，然后转到下一个列表并循环遍历其内容。

keywords = [
    ["wet","water","soaked","liquid"],
    ["cracked","broken","smashed","snapped"],
    ["dead","charge","flat","turn on"]]

for i in keywords:
    print i

Answer 1

试试这个：

>>> keywords = [
...     ["wet","water","soaked","liquid"],
...     ["cracked","broken","smashed","snapped"],
...     ["dead","charge","flat","turn on"]]
>>> for i in keywords:
...    for content in i:
...      print content

Answer 2

试试这个

   for i in keywords:
        for j in i:
            print j

Answer 3

for n in keywords:
    for i in n:
        print(i)

Answer 4

你可以使用发电机

for i in (el for sublist in keywords for el in sublist):
    print(i)

或列表理解，如果您需要保留结果并且不关心打印输出的格式

unwound = [el for sublist in keywords for el in sublist]
print(unwound)

Answer 5

我就是这样做的：

keywords = [
    ["wet","water","soaked","liquid"],
    ["cracked","broken","smashed","snapped"],
    ["dead","charge","flat","turn on"]]

for i in range(0, len(keywords) ):
    print keywords[i]
    for kw in keywords[i]:
        print kw

Answer 6

您可以使用list comprehension和join方法：

print "\n".join([i for j in keywords for i in j])

wet       
water     
soaked    
liquid    
cracked   
broken    
smashed   
snapped   
dead      
charge    
flat      
turn on

修改

@Pynchia建议使用generator代替list（IIUC）：

print "\n".join(i for j in keywords for i in j)

时间：

In [188]: %timeit "\n".join([i for j in l for i in j]) 1000000 loops, best of 3: 1.63 us per loop In [189]: %timeit "\n".join(i for j in l for i in j) 1000000 loops, best of 3: 1.97 us per loop

Answer 7

实际上，您可以像这样轻松访问列表中的元素：

XMPP

如果您需要通过循环访问元素，您可以编写如下代码：

print keywords[0][0] // output : "wet"

我如何在python中循环遍历2d数组

7 个答案: