2017-10-07 55 views
0

新手一般在此處進行編程。我有一個沒有數字鍵的空字典(在這種情況下不是= 3)。我試圖生成一個列表,這些列表將被添加到循環中的字典中,但是在每個循環中,其他鍵將被覆蓋。見我以下嘗試:試圖將列表的列表添加到for循環中的空字典中,但是在每次迭代中條目都被完全覆蓋

#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) 

這是當我檢查進展的輸出:

{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]]} 

對於上下文中,這是產生具有相同纖維元件內的一個隨機分佈的ABAQUS網格文件矩陣。

編輯:上下文


使用copy.deepcopy工作

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) 

輸出:

{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]]} 
+1

請參閱https://stackoverflow.com/q/2612802/8586227 –

+0

謝謝。 copy.deepcopy爲我工作 – Aedan

回答

1

特別是在你的情況,只是分配表達你分配至fibre_efibre[a]代替。然後爲每次迭代創建新列表。 (您根本不需要變量fibre_e

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