看着顯示器在運行MATLAB
a = 1
a =
1 2
a =
1 2 3
a =
1 2 3 4
.... (so on for 100 iterations)
在八音我可以這樣做:
>> i=1:10
i =
1 2 3 4 5 6 7 8 9 10
>> j=(1:10)'
j =
1
2
3
4
5
6
7
8
9
10
>> a=i+j
a =
2 3 4 5 6 7 8 9 10 11
3 4 5 6 7 8 9 10 11 12
4 5 6 7 8 9 10 11 12 13
5 6 7 8 9 10 11 12 13 14
6 7 8 9 10 11 12 13 14 15
7 8 9 10 11 12 13 14 15 16
8 9 10 11 12 13 14 15 16 17
9 10 11 12 13 14 15 16 17 18
10 11 12 13 14 15 16 17 18 19
11 12 13 14 15 16 17 18 19 20
這使得利用廣播,這一概念從numpy的
In [500]: i=np.arange(1,11)
In [501]: a = i[:,None] + i
In [502]: a
Out[502]:
array([[ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
[ 4, 5, 6, 7, 8, 9, 10, 11, 12, 13],
[ 5, 6, 7, 8, 9, 10, 11, 12, 13, 14],
[ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 7, 8, 9, 10, 11, 12, 13, 14, 15, 16],
[ 8, 9, 10, 11, 12, 13, 14, 15, 16, 17],
[ 9, 10, 11, 12, 13, 14, 15, 16, 17, 18],
[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
[11, 12, 13, 14, 15, 16, 17, 18, 19, 20]])
借
這是最佳實踐 - 在numpy中,我敢說MATLAB和Octave。
但是如果你必須使用迭代這樣做
In [503]: a=np.zeros((10,10),int)
In [504]: for i in range(10):
...: for j in range(10):
...: a[i,j]=i+j
具有完全成熟的Python列表進行迭代:
In [512]: alist = []
In [513]: for i in range(10):
...: sublist=[]
...: for j in range(10):
...: sublist.append(i+j)
...: alist.append(sublist)
...:
In [514]: alist
Out[514]:
[[0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
[2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
[4, 5, 6, 7, 8, 9, 10, 11, 12, 13],
[5, 6, 7, 8, 9, 10, 11, 12, 13, 14],
[6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[7, 8, 9, 10, 11, 12, 13, 14, 15, 16],
[8, 9, 10, 11, 12, 13, 14, 15, 16, 17],
[9, 10, 11, 12, 13, 14, 15, 16, 17, 18]]
In [515]: np.array(alist)
Out[515]:
array([[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9],
[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
[ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11],
[ 3, 4, 5, 6, 7, 8, 9, 10, 11, 12],
[ 4, 5, 6, 7, 8, 9, 10, 11, 12, 13],
[ 5, 6, 7, 8, 9, 10, 11, 12, 13, 14],
[ 6, 7, 8, 9, 10, 11, 12, 13, 14, 15],
[ 7, 8, 9, 10, 11, 12, 13, 14, 15, 16],
[ 8, 9, 10, 11, 12, 13, 14, 15, 16, 17],
[ 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]])
,但我可以生成alist
更緊湊與
alist=[[i+j for i in range(10)] for j in range(10)]
當您建立一個列表清單時,確保所有子列表都具有相同的leng - 否則你會回過頭來問問題。
對於循環和動態分配是一種不好的做法,在Matlab中和Python一樣。 – UpSampler
如果你必須像這樣迭代,以'a = np.zeros((10,10),dtype = int)'開頭。 – hpaulj
MATLAB的數組實際上並不是動態的。像numpy數組一樣,如果不創建新數組並且複製所有數據,就不能調整MATLAB數組的大小。這只是假裝當numpy沒有調整它們的大小時。 – TheBlackCat