我有一个很长的序列列表(假设每个长度为 16),由 0 和 1 组成。例如s = ['0100100000010111', '1100100010010101', '1100100000010000', '0111100011110111', '1111100011010111']现在我想把每一位都当作一个特征,所以我需要把它转换成 numpy 数组或 Pandas 数据帧。为了做到这一点,我需要用逗号分隔序列中存在的所有位,这对于大数据集是不可能的。所以我尝试的是生成字符串中的所有位置:slices = []for j in range(len(s[0])): slices.append((j,j+1)) print(slices)[(0, 1), (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9), (9, 10), (10, 11), (11, 12), (12, 13), (13, 14), (14, 15), (15, 16)]new = []for i in range(len(s)): seq = s[i] for j in range(len(s[i])): ## I have tried both of these LOC but couldn't figure out ## how it could be done new.append([s[slice(*slc)] for slc in slices]) new.append(s[j:j+1])print(new)预期o/p:new = [[0,1,0,0,1,0,0,0,0,0,0,1,0,1,1,1], [1,1,0,0,1,0,0,0,1,0,0,1,0,1,0,1], [1,1,0,0,1,0,0,0,0,0,0,1,0,0,0,0], [0,1,1,1,1,0,0,0,1,1,1,1,0,1,1,1], [1,1,1,1,1,0,0,0,1,1,0,1,0,1,1,1]]提前致谢!!
2 回答
猛跑小猪
TA贡献1858条经验 获得超8个赞
使用np.array构造函数和列表推导式:
np.array([list(row) for row in s], dtype=int)
array([[0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1],
[1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 1],
[1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
[0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1],
[1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1]])
jeck猫
TA贡献1909条经验 获得超7个赞
在一行中,没有for循环:
np.array(s).view('<U1').astype(int).reshape(len(s), -1)
array([[0, 1, 0, ..., 1, 1, 1],
[1, 1, 0, ..., 1, 0, 1],
[1, 1, 0, ..., 0, 0, 0],
[0, 1, 1, ..., 1, 1, 1],
[1, 1, 1, ..., 1, 1, 1]])
虽然仍然比列表理解慢一点
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