这里是编程新手。我有一本没有多少键的空字典(在本例中no=3)。我试图生成一个列表列表,该列表将在循环中添加到字典中,但在每个循环中,其他键都被覆盖。请看下面我的尝试:
#generating dictionary for all fibres with number n
fibre = {i : {}for i in range(no)}
#generating empty fibre of size n_f
fibre_e = [[[] for i in range(n_f)] for j in range(n_f)]
rand_r = [[] for a in range(no)]
rand_c = [[] for a in range(no)]
#generating fibre element from random corner elements
for a in range(no):
# choosing a random corner element
fibre[a] = fibre_e
rand_r[a] = randrange(len(all_e))
rand_c[a] = randrange(len(all_e))
for i in range(n_f):
e1_fibre=all_e[rand_r[a]][rand_c[a]]
#move to upper row
rand_r[a] += 1
for j in range(n_f):
fibre_e[i][j] = e1_fibre
e1_fibre += 1
print(fibre)
这是我检查进度时的输出:
{0: [[57, 58], [66, 67]], 1: {}, 2: {}}
{0: [[33, 34], [42, 43]], 1: [[33, 34], [42, 43]], 2: {}}
{0: [[29, 30], [38, 39]], 1: [[29, 30], [38, 39]], 2: [[29, 30], [38, 39]]}
对于上下文,这是生成一个abaqus网格文件,其中矩阵中随机分布着相同的纤维元素
你知道吗edit:context 你知道吗
使用copy.deepcopy
for a in range(no):
# choosing a random corner element
rand_r[a] = randrange(len(all_e))
rand_c[a] = randrange(len(all_e))
for i in range(n_f):
e1_fibre=all_e[rand_r[a]][rand_c[a]]
#move to upper row
rand_r[a] += 1
for j in range(n_f):
fibre_e[i][j] = e1_fibre
e1_fibre += 1
fibre[a] = copy.deepcopy(fibre_e)
print(fibre)
输出:
{0: [[68, 69], [77, 78]], 1: {}, 2: {}}
{0: [[68, 69], [77, 78]], 1: [[6, 7], [15, 16]], 2: {}}
{0: [[68, 69], [77, 78]], 1: [[6, 7], [15, 16]], 2: [[28, 29], [37, 38]]}
具体来说,只需将表达式指定给
fibre_e
到fibre[a]
。然后为每个迭代创建新的列表(您根本不需要变量fibre_e
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