Python 从具有偶然模式的列表中提取元素到元组中

Question

我有一个列表，它有这样的模式 [float, string, float string...] 但偶尔在模式中它会更改为 [float, string, float, string, string, float string.. .]。我想要做的是将列表的元素提取到格式为 (float, name, NoneType or str) 的元组中，以便稍后对其进行处理。这是一个小例子：

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，使元组看起来像这样：

(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，我已经尝试检查该类型数组中的下一个索引，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)

Answer 1

您可以使用辅助列表跟踪自上次浮动值以来看到的数组元素。每当看到float，就把已有的元素转成元组，清空辅助列表：

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)

这输出：

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

Answer 2

您可以检查“浮动”、“字符串”模式并相应地附加：

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')]

Answer 3

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)

打印：

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

Python 从具有偶然模式的列表中提取元素到元组中

Python extract elements from list with occasional pattern into a tuple

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