使用循环或列表创建变量。 动态循环不好吗？

Question

So i've looked through stackoverflow and feel that very similar questions have been asked, however I wasn't able to quite tailor those answers to my problem (sorry!).

From other threads, what I have gathered is that 1) dynamic loops are bad 2) list comprehension or dictionaries might help me?

What I want to end up with

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= ...

Sort of what I want to do

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 = .....

Any suggestions, or confusions regarding what I am trying to do here?

Thanks!

Answer 1

You have a bunch a variables that you want to assign to, where the names might look like these: grant0, funding97, grant42, grant26, funding27

So instead of having seperate variables, put the data all together into one variable that stores them like this:

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

So instead of accessing the data from the variable grant0 , you can do source['grant'][0] . As another example, instead of using funding97 , use source['funding'][97]

---

If I understand your question correctly, this may be what you want:

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]))

---

How list comprehensions work:

>>> 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]

Dictionary comprehensions work similar to list comprehensions, except you have key value pairs.

---

Step-by-step explanation of my solution using the interpreter:

>>> 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

What's wrong with "dynamic loops"? From the point of view of code reuse, your latter approach is much more elegant:

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)

I am not altogether sure what you are attempting to do. Hope this helps!