爲什麼在無限循環上這個自定義平分函數？ 。

Question

我在我最初使用平分在相當長的腳本這是它的一部分（工作完全正常，併爲意）：

portfolios = [[1], [0.9], [0.8], [0.7], [0.6]] #Fills up list to avoid "index out of range" error later on in code 
add_sharpe = [sharpe, name_a, weight_a, exchange_a, name_b, weight_b, exchange_b, name_c, weight_c, exchange_c] 
for x in portfolios: 
    if sharpe > x[0]: 
     sharpes = [i[0] for i in portfolios] 
     if name_a not in x and name_b not in x and name_c not in x: 
      print sharpe 
      print sharpes 
      print portfolios 
      print add_sharpe 
      position = reverse_bisect(sharpes, sharpe) 
      print position 
      portfolios.insert(position, add_sharpe) 

不過，現在我需要一個反向對開（降序）。謝天謝地，我發現了a really good solution to this。

的代碼創建反向對開如下：

def reverse_bisect(a, x, lo=0, hi=None): 
    if lo < 0: 
     raise ValueError('lo must be non-negative') 
    if hi is None: 
     hi = len(a) 
    while lo < hi: 
     mid = (lo+hi)//2 
     if x > a[mid]: hi = mid 
     else: lo = mid+1 
    return lo 

它非常好，當我在外面用簡單的計算測試它。但是，當我將它插入到腳本中時，它會在運行時導致腳本凍結。我不知道爲什麼會發生這種情況，因爲我使用的工作原理與bisect.bisect完全一樣。

這是什麼現在不工作：

 if name_a not in x and name_b not in x and name_c not in x: 
      position = reverse_bisect(sharpes, sharpe) 
      portfolios.insert(position, add_sharpe) 

出於某種原因，使用功能似乎遍歷portfolios.insert(position, add_sharpe)沒有結束。

輸出：

[1, 0.9, 0.8, 0.7, 0.6] #print portfolios 
[[1], [0.9], [0.8], [0.7], [0.6]] #print sharpes 
[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '] #print portfolios 
0 #print position 
1.62759369021 #print sharpe 
[1.6275936902107178, 1, 0.9, 0.8, 0.7, 0.6] #print portfolios 
[[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '], [1], [0.9], [0.8], [0.7], [0.6]] 
[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '] 
1 
1.62759369021 
[1.6275936902107178, 1.6275936902107178, 1, 0.9, 0.8, 0.7, 0.6] 
[[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '], [1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '], [1], [0.9], [0.8], [0.7], [0.6]] 
[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '] 
2 
1.62759369021 
[1.6275936902107178, 1.6275936902107178, 1.6275936902107178, 1, 0.9, 0.8, 0.7, 0.6] 
[[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '], [1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '], [1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '], [1], [0.9], [0.8], [0.7], [0.6]] 
[1.6275936902107178, "CARR'S GROUP (CARR.L)", 0.9, 'LSE ', 'Church & Dwight Co. Inc. (CHD)', 0.05, 'NYSE ', 'NCC Group plc. (NCC.L)', 0.05, 'LSE '] 
3 

Answer 1

我認爲你是插入到你迭代的列表。例如：

a = [1] 

for x in a: 
    a.insert(0, x) 
    print a 

這將讓你在一個循環中，你一直插入到1a。