我有一個(gè)很長(zhǎng)的序列列表（假設(shè)每個(gè)長(zhǎng)度為 16），由 0 和 1 組成。例如s = ['0100100000010111', '1100100010010101', '1100100000010000', '0111100011110111', '1111100011010111']現(xiàn)在我想把每一位都當(dāng)作一個(gè)特征，所以我需要把它轉(zhuǎn)換成 numpy 數(shù)組或 Pandas 數(shù)據(jù)幀。為了做到這一點(diǎn)，我需要用逗號(hào)分隔序列中存在的所有位，這對(duì)于大數(shù)據(jù)集是不可能的。所以我嘗試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)預(yù)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]]提前致謝??！
2 回答
猛跑小豬
TA貢獻(xiàn)1858條經(jīng)驗(yàn) 獲得超8個(gè)贊
使用np.array構(gòu)造函數(shù)和列表推導(dǎo)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貓
TA貢獻(xiàn)1909條經(jīng)驗(yàn) 獲得超7個(gè)贊
在一行中，沒(méi)有for循環(huán)：
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]])
雖然仍然比列表理解慢一點(diǎn)
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