如何把嵌套的python list轉成一個一維的python list?

a = [[1,2,3], [5, 2, 8], [7,8,9]], 想把a變成[1,2,3, 5, 2, 8, 7,8,9]

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python2的話

&>&>&> from compiler.ast import flatten
&>&>&> flatten(a)


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來個簡單點的

或者：

import itertools
a = [[1,2,3],[4,5,6], [7], [8,9]]
out = list(itertools.chain.from_iterable(a))


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reduce(lambda x,y:x+y, a)

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import operator
from functools import reduce
a = [[1,2,3], [4,5,6], [7,8,9]]
print(reduce(operator.add, a))


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最符合直覺的不應該是把沒用的方括弧去掉么（逃

a = eval("[%s]"%repr(l).replace("[", "").replace("]", ""))


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謝邀。

方式比較多，遞歸比較容易懂：

def expand_list(nested_list):
for item in nested_list:
if isinstance(item, (list, tuple)):
for sub_item in expand_list(item):
yield sub_item
else:
yield item

print list(expand_list([[1,2,3], [5, 2, 8], [7,8,9]]))


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來個one-line的版本，可以展開任意深度的嵌套 |??)

from collections import Iterable

flat = lambda t: [x for sub in t for x in flat(sub)] if isinstance(t, Iterable) else [t]

In [1]: flat([1,2,[3,4,[5]],[6,7]])
Out[1]: [1, 2, 3, 4, 5, 6, 7]


或者用Generator版本：

flat = lambda t: (x for sub in t for x in flat(sub)) if isinstance(t, Iterable) else iter((t,))

In [2]: flat([[7,6],[5,[4,[3]], [2,1]]])
Out[2]: & at 0x000000000410CAB0&>

In [3]: list(_)
Out[3]: [7, 6, 5, 4, 3, 2, 1]


--------------------------分-------------------割--------------------線---------------------------

如果用寬度優先展開嵌套呢？( ???)

......

先仔細觀察深度優先迭代的標準型：

def flat_depth(tree):
if not isinstance(tree, Iterable):
yield tree
else:
for sub in tree:
for x in flat_depth(sub):
yield x


思考中）那如果需要寬度優先，是不是可以讓else分支里，tree裡面的每個分支排一下隊？

諾~

def flat_breadth(tree):
if not isinstance(tree, Iterable):
yield tree
else:
for x in zipper(map(flat_breadth, tree)):
yield x


那麼問題來了 --

What &TF& is zipper？

-- zipper就是一個隊列處理器，輸入一系列迭代器，它來處理輪流迭代的循環。

from collections import deque

def zipper(itrs):
q = deque(itrs)
while q:
itr = q.popleft()
try:
yield next(itr)
q.append(itr)
except StopIteration:
pass


隨意感受一下：

b = [1,
[2, 3],
[4, [5, 6],
[7, 8]]]

# 深度優先
In [1]: list(flat_depth(b))
Out[1]: [1, 2, 3, 4, 5, 6, 7, 8]

# 寬度優先
In [2]: list(flat_breadth(b))
Out[2]: [1, 2, 4, 3, 5, 7, 6, 8]


答非所問的地方還望題主見諒( ??。)

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如何把嵌套的python list轉成一個一維的python list? - 郝小胖的回答

對字元串的支持有問題，上面的回答大多都是深度優先的版本，我來提供一個 簡單粗暴 廣度 優先展開 非遞歸 的 版本 , 只用到python內置函數，不依賴任何庫 ：

def flap(*a):
"""Breadth-First Traversal"""
a=list(a);r=[];i=0
while -1&

時間複雜度 o(n) ,最壞情況下空間複雜度

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適合字元和數字情況，並且支持多層嵌套的來一個：

def flat(l):
for k in l:
if not isinstance(k, (list, tuple)):
yield k
else:
yield from flat(k)


結果：

&>&>&> a = ["參與/A", ["北京/G", "電影/C", "學院/D"], "和/X", ["美國/G", "辛普森/F", "公司/D"], "的/B", "活動/Z", "，/Z", "由/A", ["交通/C", "銀行/D", "北京/G", "分行/D"], "與/X", "麥當勞/K", "贊助/B", "，/A", ["巴/J", "政府/D"], "和/X", ["中共中央/K", "顧問/K", "委員會/D"], "指導/B", "，/A", ["中央/C", "電視台/D"], "報道/B"]
&>&>&> list(flat(a))
["參與/A", "北京/G", "電影/C", "學院/D", "和/X", "美國/G", "辛普森/F", "公司/D", "的/B", "活動/Z", "，/Z", "由/A", "交通/C", "銀行/D", "北京/G", "分行/D", "與/X", "麥當勞/K", "贊助/B", "，/A", "巴/J", "政府/D", "和/X", "中共中央/K", "顧問/K", "委員會/D", "指導/B", "，/A", "中央/C", "電視台/D", "報道/B"]

&>&>&>list(flat([[1,[2,3]], [5, 2, [8]], [7,8,9]]))
[1, 2, 3, 5, 2, 8, 7, 8, 9]


---

&>&>&> import itertools
&>&>&> list(itertools.chain.from_iterable(a))


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func = lambda x: [y for l in x for y in func(l)] if type(x) is list else [x]
result = func(target_list)


可以展開任意維

誠如諸位所說，這種問題就是 let me google it for you 的問題

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提問之前最好Stackoverflow找找,

flatten list python 基本就可以得到答案了.

# flatten
[item for sublist in list for item in sublist]


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提問之前最好Google一下！

Python壓平嵌套列表

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《python基礎教程》

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ab = [[1,2,3], [5, 2, 8], [7,8,9]]
print([i for item in ab for i in item ])

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def fun(li,em=[]): #優雅滴寫法，包治各種嵌套
for x in li:
if type(x)!=list:
em.append(x)
else:
fun(x)
return em


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只適用於list和tuple：

def flat(l):
def iter(l):
if isinstance(l, (list, tuple)):
for k in l:
for v in iter(k):
yield v
else:
yield l
if l is None:
return None
return list(iter(l))


但是對於自包含或遞歸包含的列表誤解，會陷入死循環的。

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a = [[1,2,3], [4,5,6], [7,8,9]]

extend_a = [number for inner_list in a for number in inner_list]

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最簡單的方法，用 numpy.ravel()。

a = [[1,2,3], [5, 2, 8], [7,8,9]]

list(np.ravel(a))

得到：

[1, 2, 3, 5, 2, 8, 7, 8, 9]

---

a = [[1,2,3], [5, 2, 8], [7,8,9]]
l=[]
for m in range(0,3):
for i in a[m]:
l.append(i)
print(l)


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