我一直在寻找这个，但我找不到任何有帮助的东西。我制作一个我想要的大小的数组时遇到了一些麻烦。
到目前为止，我已使用此代码创建了一个列表列表：
n=4
def matrixA(k):
A=[]
for m in range(0,k):
row=[]
#A.append([])
for n in range(0,k):
if (n==(m+1)) or (n==(m-1)):
row.append(-deltaX/(deltaX)**2)
if (n==m):
row.append(2*deltaX/(deltaX)**2)
else:
row.append(0)
A.append(row)
return A
pprint.pprint(matrixA(n))
print len(matrixA(n))
我得到了这个输出。
[[128.0, -64.0, 0, 0, 0],
[-64.0, 0, 128.0, -64.0, 0, 0],
[0, -64.0, 0, 128.0, -64.0, 0],
[0, 0, -64.0, 0, 128.0]]
4
现在，我想让它成为一个大小（4,4）的数组。我的问题是，当我执行以下操作（将列表转换为数组并尝试对其进行整形）时：
A=numpy.array(matrixA(n))
print A
print "This is its shape:",A.shape
A.shape=(n,n)
我获得：
[[128.0, -64.0, 0, 0, 0] [-64.0, 0, 128.0, -64.0, 0, 0]
[0, -64.0, 0, 128.0, -64.0, 0] [0, 0, -64.0, 0, 128.0]]
This is its shape: (4,)
然后是错误：
ValueError: total size of new array must be unchanged
我怎么能从这里得到一个大小（4,4）的数组呢？
答案 0 :(得分：2)
欢迎来到numpy world。
似乎你想要：
NSString * key1 =@"name";
NSString * obj1 =@"Hayagreeva";
NSString * key2 =@"age";
NSNumber * obj2 =[NSNumber numberWithInt:3];
//1.Empty dictionary. @{} dictionary representation and it is not for NSMuableDictionary
NSDictionary * dict1 =[NSDictionary dictionary];
NSDictionary * dict2 =[[NSDictionary alloc] init];
NSDictionary * dict3 = @{};
//2.Dictionary with Single Object and Key
NSDictionary * dict4 =[NSDictionary dictionaryWithObject:obj1 forKey:key1];
NSDictionary * dict5 =@{key1:obj1};
//3.Dictionary with Multiple Object and Keys
//Objects and keys are sequentially terminated with nil
NSDictionary * dict6 = [NSDictionary
dictionaryWithObjectsAndKeys:obj1,key1,obj2,key2,nil];
NSDictionary * dict7 =[[NSDictionary alloc]
initWithObjectsAndKeys:obj1,key1,obj2,key2,nil] ;
//3.Dictionary with Multiple Object and Keys
NSDictionary *dict8 =[NSDictionary
dictionaryWithObjects:@[obj1,obj2]
forKeys:@[key1,key2]];
NSDictionary *dict9 =[[NSDictionary alloc]
initWithObjects:@[obj1,obj2]
forKeys:@[key1,key2]];
NSDictionary * dict10 =@{key1:obj1,key2:obj2};
//@[obj1,obj2] is the array representation. dict6,dict7,dict8 are same.
NSDictionary *dict11 =[[NSDictionary alloc]
initWithObjects:[NSArray arrayWithObjects:obj1,obj2,nil]
forKeys:[NSArray arrayWithObjects:key1,key2,nil]];
//4.Dictionary with another Dictionary
//@[obj1,obj2] is the array representation. dict6,dict7,dict8 are same.
NSDictionary *dict12=[NSDictionary dictionaryWithDictionary:dict8];
NSDictionary *dict13=[[NSDictionary alloc] initWithDictionary:dict8];
在构建列表列表时，很难考虑行和列索引。在numpy中你首先塑造，然后填充，这通常更容易。
一步一步：
array([[ 128., -64., 0., 0.],
[ -64., 128., -64., 0.],
[ 0., -64., 128., -64.],
[ 0., 0., -64., 128.]])
所以你想要的是：
In [37]: from numpy import eye,diag,ones # some useful functions
In [38]: eye(4) # identity matrix
Out[38]:
array([[ 1., 0., 0., 0.],
[ 0., 1., 0., 0.],
[ 0., 0., 1., 0.],
[ 0., 0., 0., 1.]])
In [39]: eye(4)*128
Out[39]:
array([[ 128., 0., 0., 0.],
[ 0., 128., 0., 0.],
[ 0., 0., 128., 0.],
[ 0., 0., 0., 128.]])
In [40]: ones(3)
Out[40]: array([ 1., 1., 1.])
In [41]: diag(ones(3),1) # see help(diag)
Out[41]:
array([[ 0., 1., 0., 0.],
[ 0., 0., 1., 0.],
[ 0., 0., 0., 1.],
[ 0., 0., 0., 0.]])
In [42]: diag(ones(3),1).T # transpose
Out[42]:
array([[ 0., 0., 0., 0.],
[ 1., 0., 0., 0.],
[ 0., 1., 0., 0.],
[ 0., 0., 1., 0.]])
并运行：
def arrayA(n,deltaX):
A=eye(n) # id matrix of size n
B= diag(ones(n-1),1) # just ahead
return (2*A-B-B.T)*(deltaX/deltaX**2)
对于大型矩阵，速度更快：
In [45]: arrayA(4,1/64)
Out[45]:
array([[ 128., -64., 0., 0.],
[ -64., 128., -64., 0.],
[ 0., -64., 128., -64.],
[ 0., 0., -64., 128.]])
答案 1 :(得分：0)
将嵌套列表转换为数组非常简单，而且您正确地执行了此操作：
numpy.array(matrixA(n))
它不起作用的原因是你生成的嵌套列表实际上有不规则数量的＆＃34;列＆＃34;
有问题的部分是因为您错过了elif：
if (n==(m+1)) or (n==(m-1)):
row.append(-deltaX/(deltaX)**2)
if (n==m):
row.append(2*deltaX/(deltaX)**2)
将您的循环更改为：
for m in range(0,k-1):
row=[]
#A.append([])
for n in range(0,k-1):
if (n==(m+1)) or (n==(m-1)):
row.append(-deltaX/(deltaX)**2)
elif (n==m):
row.append(2*deltaX/(deltaX)**2)
else:
row.append(0)
A.append(row)