Python将列表中的元素以偶发模式提取到元组中

QJ123123 • 3 年前 • 1568 次点击

我有一个列表,它有这样的模式[浮动,字符串,浮动字符串…]但在模式中,它偶尔会变成[float,string,float,string,string,float string…]。我想做的是将列表中的元素以(float、name、NONETYPE或STRING)的格式提取到元组中,以便以后对其进行处理。下面是一个小例子:

arr = [1150.1, 'James', 3323.1, 'Steve', 9323.1, 'John', 1233.1, 'Gary', 'criminal', 3293.1, 'Josh', 9232.1, 'Daniel', 'criminal']

我想提取列表,使元组如下所示:

(1150.1, James, NONE)

(3323.1, Steve, NONE)

(9323.1, John, NONE)

(1233.1, Gary, criminal)

(3293.1, Josh, NONE)

(9232.1, Daniel, criminal)

到目前为止,我已经尝试在数组中检查该类型的下一个索引,但它不起作用:

for index in range(len(arr)):
    if type(arr[index]) == float and type(arr[index+1]) == str:
        tup = arr[index], arr[index+1], None
        print(tup)
    elif type(arr[index]) == float and type(arr[index+1]) == str and type(arr[index+2]) == str:
        tup = arr[index], arr[index + 1], arr[index+2]
        print(tup)

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1568 次点击

文章 [ 4 ] | 最新文章 3 年前

• 1 楼

Sash Sinha 3 年前

也许你可以使用while循环:

from typing import Optional, NamedTuple


class Element(NamedTuple):
    value: float
    name: str
    profession: Optional[str] = None


def main() -> None:
    arr = [
        1150.1, 'James',
        3323.1, 'Steve',
        9323.1, 'John',
        1233.1, 'Gary', 'criminal',
        3293.1, 'Josh',
        9232.1, 'Daniel', 'criminal',
    ]
    parsed_elements = []
    i = 0
    while i <= len(arr) - 2:
        if isinstance(arr[i + 2], str):
            parsed_elements.append(Element(*arr[i:i + 3]))
            i += 3
        else:
            parsed_elements.append(Element(*arr[i:i + 2]))
            i += 2
    print('\n'.join(str(e) for e in parsed_elements))


if __name__ == '__main__':
    main()

输出:

Element(value=1150.1, name='James', profession=None)
Element(value=3323.1, name='Steve', profession=None)
Element(value=9323.1, name='John', profession=None)
Element(value=1233.1, name='Gary', profession='criminal')
Element(value=3293.1, name='Josh', profession=None)
Element(value=9232.1, name='Daniel', profession='criminal')

• 2 楼

Andrej Kesely 3 年前

另一个解决方案:

from itertools import groupby

g1 = (g for v, g in groupby(arr, type) if v is float)
g2 = (g for v, g in groupby(arr, type) if v is str)

out = [(next(a), *[*b, None][:2]) for a, b in zip(g1, g2)]
print(out)

印刷品:

[
    (1150.1, "James", None),
    (3323.1, "Steve", None),
    (9323.1, "John", None),
    (1233.1, "Gary", "criminal"),
    (3293.1, "Josh", None),
    (9232.1, "Daniel", "criminal"),
]

• 3 楼

not_speshal 3 年前

您可以检查“float”、“string”模式,并相应地追加:

output = list()
for i, element in enumerate(arr):
    if isinstance(element, float) and isinstance(arr[i+1], str):
        if isinstance(arr[i+2], str):
            t = tuple(arr[i:i+3])
        else:
            t = tuple(arr[i:i+2]+["NONE"])
        output.append(t)

>>> output
[(1150.1, 'James', 'NONE'),
 (3323.1, 'Steve', 'NONE'),
 (9323.1, 'John', 'NONE'),
 (1233.1, 'Gary', 'criminal'),
 (3293.1, 'Josh', 'NONE'),
 (9232.1, 'Daniel', 'criminal')]

• 4 楼

BrokenBenchmark 3 年前

可以使用辅助列表跟踪自上次浮点值以来看到的数组元素。每当看到浮点时,将现有元素转换为元组并清除辅助列表:

result = []
items = []

for item in arr:
    if isinstance(item, float) and items:
        if len(items) < 3:
            items.append(None)
        result.append(tuple(items))
        items = [item]
    else:
        items.append(item)
    
result.append(tuple(items))

print(result)

这将产生:

[
 (1150.1, 'James', None), (3323.1, 'Steve', None),
 (9323.1, 'John', None), (1233.1, 'Gary', 'criminal'),
 (3293.1, 'Josh', None), (9232.1, 'Daniel', 'criminal')
]

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