怎样把python 二维元组转成一个一维的python 列表?

[Logix5000]

a = [(4,2,3), (5, 9, 1), (7,8,9)]
想把a变成["4,2,3", "5, 9, 1", "7,8,9",]

先谢谢大家了。

[Js丶]

代码如下：

a = [(4, 2, 3), (5,9,1), (7,8,9)]
i=0
while i<3:
	a[i]=str(a[i])[1:3*3-1]
	i=i+1
print (a[0:3])

输出结果为： ['4, 2, 3', '5, 9, 1', '7, 8, 9'] 感觉你的描述有点问题，a = [(4,2,3), (5, 9, 1), (7,8,9)]应该是一个列表，只不过列表里的元素是元组。 所以问题也就变成了 ：把一个列表中的各元素，从元组变成字符串。 方法即为上面代码。

[AwaiNG6]

a = [(4,2,3), (5, 9, 1), (7,8,9)] #数据 b = [] #容器b列表 for each in a: #循环数据内的元组 for each in each: #循环元组内对象 b.append(each) #追加到容器b b #打印b -----------------输出如下-------------------- a = [(4,2,3), (5, 9, 1), (7,8,9)] b = [] for each in a: for each in each: b.append(each) b Out[6]: [4, 2, 3, 5, 9, 1, 7, 8, 9]

[混沌鳄鱼]

引用 2 楼 qq_40094365 的回复:

a = [(4,2,3), (5, 9, 1), (7,8,9)] #数据 b = [] #容器b列表 for each in a: #循环数据内的元组 for each in each: #循环元组内对象 b.append(each) #追加到容器b b #打印b -----------------输出如下-------------------- a = [(4,2,3), (5, 9, 1), (7,8,9)] b = [] for each in a: for each in each: b.append(each) b Out[6]: [4, 2, 3, 5, 9, 1, 7, 8, 9]

不要自己造轮子呀 你在用的是python 不是C也不是java

>>> a = [(4,2,3), (5, 9, 1), (7,8,9)]
>>> from itertools import chain
>>> list(chain.from_iterable(a))
[4, 2, 3, 5, 9, 1, 7, 8, 9]
>>> from tkinter import _flatten   # python2.7也可以from compiler.ast import flatten
>>> _flatten(a)
(4, 2, 3, 5, 9, 1, 7, 8, 9)

>>> [','.join(map(str,t)) for t in a]
['4,2,3', '5,9,1', '7,8,9']
>>> from itertools import starmap
>>> list(starmap('{},{},{}'.format,a))
['4,2,3', '5,9,1', '7,8,9']

[oyljerry]

用列表推导

In [3]: a = [(4,2,3), (5, 9, 1), (7,8,9)]

In [8]: b = [[str(y) for y in x] for x in a]

In [9]: b
Out[9]: [['4', '2', '3'], ['5', '9', '1'], ['7', '8', '9']]

In [10]: c = [",".join(z) for z in b]

In [11]: c
Out[11]: ['4,2,3', '5,9,1', '7,8,9']