使用循環或列表創建變量。動態循環不好？

Question

所以我查看了一下stackoverflow，覺得非常類似的問題已經被問到，但是我無法完全解決這些問題的答案（對不起！）。

從其他線程，我收集的是， 1）動態循環不好 2）列表理解或字典可能幫助我？

我想

grant1 = median(df.query('income < 25000')['grant'] 
grant2 = median(df.query('income > 25000 and income < 50000')['grant'] 
grant3 = median(df.query('income > 50000 and income < 75000')['grant'] 
grant4 = ... 
grant5 = ... 

funding1 = median(df.query('income < 25000')['funding'] 
funding2 = median(df.query('income > 25000 and income < 50000')['funding'] 
funding3 = median(df.query('income > 50000 and income < 75000')['funding'] 
funding4 = ... 
funding5= ... 

排序的我想要做什麼方面，我想在這裏做

source = ['grant', 'funding'] 
for x in source: 
    x1 = median(df.query('income < 25000')['x'] 
    x2 = median(df.query('income > 25000 and income < 50000')['x'] 
    x3 = .... 
    x4 = ..... 

任何建議或混亂結束了？

謝謝！

Answer 1

你有一堆要分配到，這裏的名字看起來像這樣一個變量：grant0，funding97，grant42，grant26，funding27

因此，而不是具有單獨的變量，把數據放在一起爲一個變量存儲他們這樣的：

source = {'grant': [], 'funding': []'} 

因此，而不是從變量grant0訪問數據，你可以做source['grant'][0]。再舉一個例子，而不是使用funding97，使用source['funding'][97]

---

如果我正確理解你的問題，這可能是你想要什麼：

income_brackets = [(0, 25000), (25000, 50000)] 
source = {'grant': [], 'funding': []} 

#Hard to understand one-liner 
#source = {key: [median(df.query('income > {} and income < {}'.format(lower, upper))[key]) for lower, upper in income_brackets] for key in source} 

#Easy to understand three-liner 
for lower, upper in income_brackets: 
    for key in source: 
     source[key].append(median(df.query('income > {} and income < {}'.format(lower, upper))[key])) 

---

如何list解析的工作：

>>> foo = [] 
>>> for i in range(10): 
     foo.append(i) 
>>> foo 
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9] 
>>> bar = [i for i in range(10)] 
>>> bar 
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9] 

字典理解工作類似於列表comprehensio ns，除了你有鍵值對。

---

步驟一步解釋使用解釋我的解決方案：

>>> income_brackets = [(0, 25000), (25000, 50000)] 
>>> source = {'grant': [], 'funding': []} 
>>> for lower, upper in income_brackets: 
     print(lower, upper) 
0 25000 
25000 50000 
>>> for lower, upper in income_brackets: 
     for key in source: 
      print(lower, upper, key) 
0 25000 grant 
0 25000 funding 
25000 50000 grant 
25000 50000 funding 
>>> for lower, upper in income_brackets: 
     for key in source: 
      source[key].append(5) 
>>> source 
{'grant': [5, 5], 'funding': [5, 5]} 
>>> source['grant'][0] 
5 
>>> source['grant'][1] 
5 
>>> source['funding'][0] 
5 
>>> source['funding'][1] 
5 

Answer 2

出了什麼問題「的動態循環」？但從代碼重用的時候，你的後一種方法更優雅：

source = ['grant', 'funding'] 
result = [] 

for x in source: 
    r = [] 
    r.append(median(df.query('income < 25000')[x]) 
    r.append(median(df.query('income > 25000 and income < 50000')[x]) 
    r.append(median(df.query('income > 50000 and income < 75000')[x]) 
    # ... 
    result.append(r) 

我並不完全相信你正在試圖做的事。 希望這有助於！