如何在python中连接map的值

我有一张像这样的地图：

0,15
1,14
2,0
3,1
4,12
5,6
6,4
7,2
8,8
9,13
10,3
11,7
12,9
13,10
14,11
15,5

打印我正在做的事情

def PrintValuesArray(su_array):
    for each in su_array:
        print ",".join(map(str, each))

但是，我只希望值和逗号分开，如下所示：

我试过了

def PrintSuffixArray(su_array):
    for key, value in su_array:
        print ",".join(map(str, value))

但是得到

打印“，” 支持迭代

而且

def PrintSuffixArray(su_array):
    for key, value in su_array:
        print ",".join(map(str, value in su_array))

打印“，”.join（映射（str，su_数组中的值））类型错误：参数2到 map（）必须支持迭代

如何打印结果

15, 14, 0, 1, 12, 6, 4, 2, 8, 13, 3, 7, 9, 10, 11, 5

试试这个：

print ", ".join(str(y) for x, y in su_array)

大概是这样的：

>>> su_array = [(0, 15), (1, 14), (2, 0), (3, 1), (4, 12), (5, 6), (6, 4), (7, 2), (8, 8), (9, 13), (10, 3), (11, 7), (12, 9), (13, 10), (14, 11), (15, 5)]
>>> ', '.join(str(b) for a, b in  su_array)
'15, 14, 0, 1, 12, 6, 4, 2, 8, 13, 3, 7, 9, 10, 11, 5'

---

首先，遍历列表，只获取第二个元素（可以用于速记）。然后使用

“，”。连接（列表）

以获得所需的输出，请参阅：

或者如果它们是元组：

>>> original = """0,15
... 1,14
... 2,0
... 3,1
... 4,12
... 5,6
... 6,4
... 7,2
... 8,8
... 9,13
... 10,3
... 11,7
... 12,9
... 13,10
... 14,11
... 15,5"""
>>>
>>> original_tuples = [tuple(i.split(",")) for i in original.split("\n")]
>>> original_tuples
[('0', '15'), ('1', '14'), ('2', '0'), ('3', '1'), ('4', '12'), ('5', '6'), ('6', '4'), ('7', '2'), ('8', '8'), ('9', '13'), ('10', '3'), ('11', '7'), ('12', '9'), ('13', '10'), ('14', '11'), ('15', '5')]
>>> ",".join(map(str,[j for i,j in original_tuples]))
'15,14,0,1,12,6,4,2,8,13,3,7,9,10,11,5'

或者，您可以跳过

映射（str，list）

并将元素转换为列表中的字符串

>>> ",".join(str(j) for i,j in original_tuples)
'15,14,0,1,12,6,4,2,8,13,3,7,9,10,11,5'

>>> ",".join(str(j) for i,j in original_tuples)
'15,14,0,1,12,6,4,2,8,13,3,7,9,10,11,5'

>>> a = """0,15
... 1,14
... 2,0
... 3,1
... 4,12
... 5,6
... 6,4
... 7,2
... 8,8
... 9,13
... 10,3
... 11,7
... 12,9
... 13,10
... 14,11
... 15,5
... """
>>> map(lambda x: x.split(',')[1], a.split())
['15', '14', '0', '1', '12', '6', '4', '2', '8', '13', '3', '7', '9', '10', '11', '5']