在python中迭代向数组追加列
假设我有以下三个列表:在python中迭代向数组追加列,python,arrays,numpy,matrix,Python,Arrays,Numpy,Matrix,假设我有以下三个列表: calc_points=np.asarray( [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75,
calc_points=np.asarray(
[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47,
49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81,
83, 85, 87, 89, 91, 93, 95, 97, 99])
out=[c+1 for c in calc_points]
inout=[c+3 for c in calc_points]
我想把它们加入一个矩阵,第一列是calc_points,然后是inout,然后是out,然后是inout和out。因此,第一列只存在一次,而另两列重复5次
我试着这样做:
temp=[np.c_[calc_points,inout,out] for i in range(5)]
但它并不像想象的那样有效。而不是
计算点| inout | out | inout | out
它产生
计算点|输入|输出
计算点|输入|输出
首先使用列表理解构造列,然后连接它们:
np.stack([calc_points]+[col for _ in range(5) for col in [calc_points+3, calc_points+1]], axis=-1)
#array([[ 0, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1],
# [ 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2],
# [ 2, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3],
# [ 3, 6, 4, 6, 4, 6, 4, 6, 4, 6, 4],
# [ 4, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5],
# [ 5, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6],
# ...
首先使用列表理解构造列,然后连接它们:
np.stack([calc_points]+[col for _ in range(5) for col in [calc_points+3, calc_points+1]], axis=-1)
#array([[ 0, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1],
# [ 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2],
# [ 2, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3],
# [ 3, 6, 4, 6, 4, 6, 4, 6, 4, 6, 4],
# [ 4, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5],
# [ 5, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6],
# ...
列表理解度太低了一级。但是,您可以简单地在下标中使用列表理解:
np.c_[(calc_points,)+(inout,out)*5]
其中:
>>> np.c_[(calc_points,)+(inout,out)*5]
array([[ 0, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1],
[ 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2],
[ 2, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3],
[ 3, 6, 4, 6, 4, 6, 4, 6, 4, 6, 4],
[ 4, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5],
[ 5, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6],
[ 6, 9, 7, 9, 7, 9, 7, 9, 7, 9, 7],
[ 7, 10, 8, 10, 8, 10, 8, 10, 8, 10, 8],
[ 8, 11, 9, 11, 9, 11, 9, 11, 9, 11, 9],
[ 9, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10],
[ 10, 13, 11, 13, 11, 13, 11, 13, 11, 13, 11],
[ 11, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12],
[ 12, 15, 13, 15, 13, 15, 13, 15, 13, 15, 13],
[ 13, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14],
等等列表理解度太低了一级。但是,您可以简单地在下标中使用列表理解:
np.c_[(calc_points,)+(inout,out)*5]
其中:
>>> np.c_[(calc_points,)+(inout,out)*5]
array([[ 0, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1],
[ 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2],
[ 2, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3],
[ 3, 6, 4, 6, 4, 6, 4, 6, 4, 6, 4],
[ 4, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5],
[ 5, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6],
[ 6, 9, 7, 9, 7, 9, 7, 9, 7, 9, 7],
[ 7, 10, 8, 10, 8, 10, 8, 10, 8, 10, 8],
[ 8, 11, 9, 11, 9, 11, 9, 11, 9, 11, 9],
[ 9, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10],
[ 10, 13, 11, 13, 11, 13, 11, 13, 11, 13, 11],
[ 11, 14, 12, 14, 12, 14, 12, 14, 12, 14, 12],
[ 12, 15, 13, 15, 13, 15, 13, 15, 13, 15, 13],
[ 13, 16, 14, 16, 14, 16, 14, 16, 14, 16, 14],
依此类推这是另一个不涉及循环的答案
# make calc_points a column vector
In [49]: calc_points[:, np.newaxis]
# make array from the list repetitions
In [50]: np.array((inout, out) * 5).T
# concatenate all of them using np.hstack
In [51]: np.hstack((calc_points[:, np.newaxis], np.array([inout, out] * 5).T))
Out[51]:
array([[ 0, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1],
[ 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2],
[ 2, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3],
[ 3, 6, 4, 6, 4, 6, 4, 6, 4, 6, 4],
[ 4, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5],
[ 5, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6],
[ 6, 9, 7, 9, 7, 9, 7, 9, 7, 9, 7],
[ 7, 10, 8, 10, 8, 10, 8, 10, 8, 10, 8],
[ 8, 11, 9, 11, 9, 11, 9, 11, 9, 11, 9],
[ 9, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10],
[ 10, 13, 11, 13, 11, 13, 11, 13, 11, 13, 11],
.....
.....
[ 99, 102, 100, 102, 100, 102, 100, 102, 100, 102, 100]])
效率:按降序排列
# interestingly list comprehension seems to be running like a war horse.
In [52]: %timeit np.stack([calc_points]+[col for _ in range(5) for col in [calc_points+3, calc_points+1]], axis=-1)
10000 loops, best of 3: 81.9 µs per loop
# almost 5x faster than using `np.c_`
In [53]: %timeit np.hstack((calc_points[:, np.newaxis], np.array((inout, out) * 5).T))
10000 loops, best of 3: 98.6 µs per loop
In [54]: %timeit np.c_[(calc_points,)+(inout,out)*5]
1000 loops, best of 3: 458 µs per loop
还有一个不涉及循环的答案
# make calc_points a column vector
In [49]: calc_points[:, np.newaxis]
# make array from the list repetitions
In [50]: np.array((inout, out) * 5).T
# concatenate all of them using np.hstack
In [51]: np.hstack((calc_points[:, np.newaxis], np.array([inout, out] * 5).T))
Out[51]:
array([[ 0, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1],
[ 1, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2],
[ 2, 5, 3, 5, 3, 5, 3, 5, 3, 5, 3],
[ 3, 6, 4, 6, 4, 6, 4, 6, 4, 6, 4],
[ 4, 7, 5, 7, 5, 7, 5, 7, 5, 7, 5],
[ 5, 8, 6, 8, 6, 8, 6, 8, 6, 8, 6],
[ 6, 9, 7, 9, 7, 9, 7, 9, 7, 9, 7],
[ 7, 10, 8, 10, 8, 10, 8, 10, 8, 10, 8],
[ 8, 11, 9, 11, 9, 11, 9, 11, 9, 11, 9],
[ 9, 12, 10, 12, 10, 12, 10, 12, 10, 12, 10],
[ 10, 13, 11, 13, 11, 13, 11, 13, 11, 13, 11],
.....
.....
[ 99, 102, 100, 102, 100, 102, 100, 102, 100, 102, 100]])
效率:按降序排列
# interestingly list comprehension seems to be running like a war horse.
In [52]: %timeit np.stack([calc_points]+[col for _ in range(5) for col in [calc_points+3, calc_points+1]], axis=-1)
10000 loops, best of 3: 81.9 µs per loop
# almost 5x faster than using `np.c_`
In [53]: %timeit np.hstack((calc_points[:, np.newaxis], np.array((inout, out) * 5).T))
10000 loops, best of 3: 98.6 µs per loop
In [54]: %timeit np.c_[(calc_points,)+(inout,out)*5]
1000 loops, best of 3: 458 µs per loop
如果in/out all只为点添加一个值,则甚至不需要连接。只需使用“外部”加法:points[:,None]+np.array[0]+[1,3]*5如果in/out all只向点添加一个值,则甚至不需要连接。只需使用“外部”加法:points[:,None]+np.array[0]+[1,3]*5在5之后有一个额外的方括号:在5之后有一个额外的方括号: