Question

我有一个这样的列表

['1133', '1300', '1418', '1443', '1473', '1600', '1601', '1988', '1990', '1991', '1992', '1993', '1994', '1995', '1996', '1997', '1998', '2000', '2003', '2004', '2005', '2006', '2007', '2008', '2009', '2010', '2012', '2013', '2014', '2015', '2153', '2600', '3000', '3714', '3785', '3896', '3995', '4001', '4436', '5094', '5346', '8000']

如何找到包含连续的数字字符串（例如：'2012'，'2013'，'2014'，'2015'）的子组？

谢谢！

Answer 1

尝试这样的事情：

    def create_model():
        lay1=Conv2D(2,kernel_size=(1,1),activation='relu',padding='same')(inputs)
        lay1=Conv2D(2,kernel_size=(7,7),strides=(2,2),activation='relu',padding='same')(lay1)
        lay1=Conv2D(2,kernel_size=(1,1),activation='relu',padding='same')(lay1)
        lay1=MaxPooling2D(pool_size=(7,7),strides=(2,2),padding='same')(lay1)
        lay2=Conv2D(4,kernel_size=(1,1),activation='relu',padding='same')(inputs)
        lay2=Conv2D(4,kernel_size=(7,7),strides=(2,2),activation='relu',padding='same')(lay2)
        lay2=Conv2D(4,kernel_size=(1,1),activation='relu',padding='same')(lay2)
        lay2=MaxPooling2D(pool_size=(7,7),strides=(2,2),padding='same')(lay2)
        lay3=Conv2D(6,kernel_size=(1,1),activation='relu',padding='same')(inputs)
        lay3=Conv2D(6,kernel_size=(7,7),strides=(2,2),activation='relu',padding='same')(lay3)
        lay3=Conv2D(6,kernel_size=(1,1),activation='relu',padding='same')(lay3)
        lay3=MaxPooling2D(pool_size=(7,7),strides=(2,2),padding='same')(lay3)
        fc=concatenate([lay1,lay2,lay3])
        fc=Flatten()(fc)
        fc=Dense(10,activation='relu')(fc)
        outputs=Dense(10,activation='softmax')(fc)
        model=Model(input=inputs,output=outputs)
        model.compile(loss=keras.losses.categorical_crossentropy,
          optimizer=keras.optimizers.Adadelta(),
          metrics=['accuracy'])
        return model

输出：

l = ['1133', '1300', '1418', '1443', '1473', '1600', '1601', '1988', '1990', '1991', '1992', '1993', '1994', '1995', '1996', '1997', '1998', '2000', '2003', '2004', '2005', '2006', '2007', '2008', '2009', '2010', '2012', '2013', '2014', '2015', '2153', '2600', '3000', '3714', '3785', '3896', '3995', '4001', '4436', '5094', '5346', '8000']

size = len(l)
i = 0
while i < size:
   j = i
   while j + 1 < size and int(l[j]) + 1 == int(l[j + 1]):
      j += 1
   if j != i:
       print (l[i:j+1])
   i = j + 1

Answer 2

如果您将指定的所有元素都放在同一个result列表中，这是一种实现方法

a = ['1133', '1300', '1418', '1443', '1473', '1600', '1601', '1988', '1990', '1991', '1992', '1993', '1994', '1995',
     '1996', '1997', '1998', '2000', '2003', '2004', '2005', '2006', '2007', '2008', '2009', '2010', '2012', '2013',
     '2014', '2015', '2153', '2600', '3000', '3714', '3785', '3896', '3995', '4001', '4436', '5094', '5346', '8000']
result = []
for i, x in enumerate(a):
    previous_e = -9999 if i == 0 else int(a[i - 1])
    next_e = -9999 if i == len(a) - 1 else int(a[i + 1])
    if int(x) == previous_e + 1 or int(x) == next_e - 1:
        result.append(x)

print(result)

['1600'，'1601'，'1991'，'1992'，'1993'，'1994'，'1995'，'1996'，'1997'，'1998'，'2004'，'2005 '，'2006'，'2007'，'2008'，'2009'，'2010'，'2013'，'2014'，'2015']

Answer 3

如果将列表转换为int列表。

mylist = list(map(int, mylist))

然后，您可以应用写成here的答案。但是，如果您使用的是python 3，请不要忘记将其转换为python 3语法。

在Python列表中查找表示数字的连续字符串

3 个答案: