问题描述
我在 Python 中有一个非常非常奇怪的简单问题.
I have a simple problem in Python that is very very strange.
def estExt(matriz,erro):
# (1) Determinar o vector X das soluções
print ("Matrix after:");
print(matriz);
aux=matriz;
x=solucoes(aux); # IF aux is a copy of matrix, why the matrix is changed??
print ("Matrix before: ");
print(matriz)
...
如下所示,尽管 aux
被函数 solucoes()改变,但矩阵
matriz
被改变了代码>.
As you see below, the matrix matriz
is changed in spite of the fact that aux
is the one being changed by the function solucoes()
.
之前的矩阵:[[7, 8, 9, 24], [8, 9, 10, 27], [9, 10, 8, 27]]
矩阵之后:
[[[7, 8, 9, 24], [0.0, -0.14285714285714235, -0.2857142857142847, -0.42857142857142705],[0.0, 0.0, -3.0, -3.0000000000000018]]
推荐答案
行
aux=matriz;
不复制matriz
,它只是创建一个名为aux
的对matriz
的新引用.你可能想要
Does not make a copy of matriz
, it merely creates a new reference to matriz
named aux
. You probably want
aux=matriz[:]
假设 matriz
是一个简单的数据结构,这将创建一个副本.如果它更复杂,你应该使用 copy.deepcopy
Which will make a copy, assuming matriz
is a simple data structure. If it is more complex, you should probably use copy.deepcopy
aux = copy.deepcopy(matriz)
顺便说一句,您不需要在每个语句后使用分号,python 不会将它们用作 EOL 标记.
As an aside, you don't need semi-colons after each statement, python doesn't use them as EOL markers.
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