Question

我有一个这样的元组列表：

[('1001794', 'Us/Eastern', '2', '1', '3', '4', '5')
('1001832', 'Us/Central', '2', '3', '4', '4', '5')
('1001848', 'Us/Central', '2', '4', '5', '4', '5')
('1001855', 'Us/Central', '2', '1', '4', '4', '5')
('1001899', 'Us/Central', '2', '1', '4', '3', '5')
('1001914', 'Us/Pacific', '1', '4', '2', '4', '5')
('1001971', 'Us/Pacific', '3', '4', '2', '3', '5')
('1002021', 'Us/Eastern', '2', '1', '4', '4', '5')
('1002026', 'Us/Central', '2', '1', '4', '4', '2')
('1002028', 'Us/Eastern', '2', '1', '4', '4', '5')
('1002041', 'Us/Eastern', '2', '4', '3', '5', '4')]

我需要将其中的所有数字转换为数字。我不介意如果必要的话我是否必须做6种不同的列表理解，但是我仍然不能将自己的头围在列表理解上，以至于不能完全从零开始。我尝试这样做：

[x = (int(x[0]),x[1],int(x[2]),int(x[3]),int(x[4]),int(x[5]),int(x[6])) for x in tlist]

但是它告诉我语法无效，我不确定为什么吗？

Answer 1

使用str.isdigit：

[tuple(int(y) if y.isdigit() else y for y in x) for x in tup]
# [(1001794, 'Us/Eastern', 2, 1, 3, 4, 5), ...]

这是假设您字符串中的数字是整数。 tup是您的数据结构。

要处理负整数，请使用封装try-except块的函数（有关EAFP principle的信息）：

def try_convert(s):
    try:
        return int(s)
    except ValueError:
        return s

[tuple(try_convert(y) for y in x) for x in tup]
# [(1001794, 'Us/Eastern', 2, 1, 3, 4, 5), ...]

要处理int和float，可以使用嵌套块：

def try_convert(s):
    try:
        return int(s)
    except ValueError:
        try:
            return float(s)
        except ValueError:
            return s

Answer 2

这对我有用：

y = [('1001794', 'Us/Eastern', '2', '1', '3', '4', '5'),
     ('1002041', 'Us/Eastern', '2', '4', '3', '5', '4')]
x = [(int(x[0]),x[1],int(x[2]),int(x[3]),int(x[4]),int(x[5]),int(x[6])) for x in y]
print(x)
[(1001794, 'Us/Eastern', 2, 1, 3, 4, 5), (1002041, 'Us/Eastern', 2, 4, 3, 5, 4)]

您只需要x =之外的[。

Answer 3

如果您知道ints在哪里，则可以使用Extended Iterable Unpacking来简化列表理解，以尝试在哪里使用：

[(int(x), y, *map(int,v)) for x, y, *v in l]

[(1001794, 'Us/Eastern', 2, 1, 3, 4, 5),
 (1001832, 'Us/Central', 2, 3, 4, 4, 5),
 (1001848, 'Us/Central', 2, 4, 5, 4, 5),
 (1001855, 'Us/Central', 2, 1, 4, 4, 5),
 (1001899, 'Us/Central', 2, 1, 4, 3, 5),
 (1001914, 'Us/Pacific', 1, 4, 2, 4, 5),
 (1001971, 'Us/Pacific', 3, 4, 2, 3, 5),
 (1002021, 'Us/Eastern', 2, 1, 4, 4, 5),
 (1002026, 'Us/Central', 2, 1, 4, 4, 2),
 (1002028, 'Us/Eastern', 2, 1, 4, 4, 5),
 (1002041, 'Us/Eastern', 2, 4, 3, 5, 4)]

Answer 4

这是您寻找的答案吗？

list = []
newList = []

list = [
        ('1001794', 'Us/Eastern', '2', '1', '3', '4', '5'),
        ('1001832', 'Us/Central', '2', '3', '4', '4', '5'),
        ('1001848', 'Us/Central', '2', '4', '5', '4', '5'),
        ('1001855', 'Us/Central', '2', '1', '4', '4', '5'),
        ('1001899', 'Us/Central', '2', '1', '4', '3', '5'),
        ('1001914', 'Us/Pacific', '1', '4', '2', '4', '5'),
        ('1001971', 'Us/Pacific', '3', '4', '2', '3', '5'),
        ('1002021', 'Us/Eastern', '2', '1', '4', '4', '5'),
        ('1002026', 'Us/Central', '2', '1', '4', '4', '2'),
        ('1002028', 'Us/Eastern', '2', '1', '4', '4', '5'),
        ('1002041', 'Us/Eastern', '2', '4', '3', '5', '4')
        ]

newList = [[int(x[0]), x[1], int(x[2]), int(x[3]), int(x[4]), int(x[5]), int(x[6])] for x in list]

print(newList)

安全地将字符串强制转换为元组列表中的整数

4 个答案: