如何用Python将一个数组分割成元素间隔为1的多个子数组
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51,
52, 69, 72, 73, 74, 75, 76, 77, 86, 97, 98, 100, 101,
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114,
115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 129,
130, 131, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152,
153, 154, 155, 156, 157, 158, 159, 160, 197, 198, 199, 200, 201,
202, 203, 204, 206, 207, 208, 209, 210, 216, 217, 218, 219, 220,
221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233,
234, 235, 236, 237, 239, 240, 241, 242, 243, 244, 259, 271, 272,
273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285,
286, 287, 288, 289, 290, 291, 292, 293, 294])
如何用Python将一个数组分割成元素间隔为1的多个子数组
我的笨拙方法:
arr = array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51,
52, 69, 72, 73, 74, 75, 76, 77, 86, 97, 98, 100, 101,
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114,
115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 129,
130, 131, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152,
153, 154, 155, 156, 157, 158, 159, 160, 197, 198, 199, 200, 201,
202, 203, 204, 206, 207, 208, 209, 210, 216, 217, 218, 219, 220,
221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233,
234, 235, 236, 237, 239, 240, 241, 242, 243, 244, 259, 271, 272,
273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285,
286, 287, 288, 289, 290, 291, 292, 293, 294])
brr = np.append(arr,arr[-1] + 1)
crr = np.where(brr[1:] - arr != 1)
drr = []
k = 0
for i in crr:
devi.append(arr[k:i])
k = i
devi.append(arr[i:])
我来给你一个直接的解决方案。这个问题其实就是把一个数组按顺序切成多个子数组,每个子数组的元素在原数组中是连续的。
def split_array_by_step(arr, step=1):
"""
将数组分割成元素间隔为指定步长的多个子数组
参数:
arr: 输入数组
step: 元素间隔,默认为1
返回:
分割后的子数组列表
"""
if not arr:
return []
result = []
current_subarray = [arr[0]]
for i in range(1, len(arr)):
# 检查当前元素与前一个元素的差值是否为指定步长
if arr[i] - arr[i-1] == step:
current_subarray.append(arr[i])
else:
# 步长不符合要求,开始新的子数组
result.append(current_subarray)
current_subarray = [arr[i]]
# 添加最后一个子数组
result.append(current_subarray)
return result
# 示例用法
if __name__ == "__main__":
# 测试数据
test_array = [1, 2, 3, 5, 6, 7, 10, 11, 12, 15]
# 分割数组
subarrays = split_array_by_step(test_array, step=1)
print("原始数组:", test_array)
print("分割结果:")
for i, sub in enumerate(subarrays):
print(f" 子数组 {i+1}: {sub}")
# 输出:
# 原始数组: [1, 2, 3, 5, 6, 7, 10, 11, 12, 15]
# 分割结果:
# 子数组 1: [1, 2, 3]
# 子数组 2: [5, 6, 7]
# 子数组 3: [10, 11, 12]
# 子数组 4: [15]
这个函数的工作原理很简单:遍历数组,检查相邻元素的差值是否为1(或指定的步长)。如果是,就把当前元素添加到当前子数组中;如果不是,就把当前子数组保存到结果里,然后开始一个新的子数组。
如果你想要更简洁的写法,也可以用列表推导式:
def split_array_compact(arr, step=1):
if not arr:
return []
result = []
start = 0
for i in range(1, len(arr)):
if arr[i] - arr[i-1] != step:
result.append(arr[start:i])
start = i
result.append(arr[start:])
return result
两种方法都能用,看你喜欢哪种风格。
总结:遍历数组,按差值分组就行。
看不懂你的要求,代码也执行不了。
[::2 ]?看不懂代码+1
不能修改帖子,只能在这里修改有错的代码:
arr = array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,
13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51,
52, 69, 72, 73, 74, 75, 76, 77, 86, 97, 98, 100, 101,
102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114,
115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 129,
130, 131, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152,
153, 154, 155, 156, 157, 158, 159, 160, 197, 198, 199, 200, 201,
202, 203, 204, 206, 207, 208, 209, 210, 216, 217, 218, 219, 220,
221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232, 233,
234, 235, 236, 237, 239, 240, 241, 242, 243, 244, 259, 271, 272,
273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285,
286, 287, 288, 289, 290, 291, 292, 293, 294])
brr = np.append(arr,arr[-1] + 1) #使 brr 为 array([ 0, 1, 2, …, 293, 294, 295])
crr = np.where(brr[1:] - arr != 1)
drr = []
k = 0
for i in crr:
drr.append(arr[k:i])
k = i
drr.append(arr[i:])
你贴示例数组,是不是贴十来个元素就可以了,至于贴这么长的么
随意到底是想干嘛都没说清楚?
就是数组[ 0, 1, 2, 3, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 19, 23, 24, 27, 28, 29, 30]分割成[ 0, 1, 2, 3][6, 7, 8, 9, 10][12][14, 15, 16, 17, 18, 19][23, 24][27, 28, 29, 30]
你这分割的长度是依照什么
早说这个不就行了吗,是把大数组分割成元素连续的一个个数组吧?你大数组一定是升序的吗?
无序的话可以用双向头尾哈希表,有序的话直接遍历搜索 arr[i] - arr[i-1] != 1 就行了。
In [1]: a = np.array([ 0, 1, 2, 3, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 19, 23, 24, 27, 28, 29, 30])
In [2]: np.split(a,np.argwhere(np.diff(a-np.arange(a.shape[0]))!=0).flatten()+1)
Out[2]:
[array([0, 1, 2, 3]),
array([ 6, 7, 8, 9, 10]),
array([12]),
array([14, 15, 16, 17, 18, 19]),
array([23, 24]),
array([27, 28, 29, 30])]
谢谢帮助。
谢谢帮助。




