Creating a List with 3 Conditions in Python

Question

The code below is a simple modification of the code that reads a page with Selenium and stores the address in a variable according to the condition of the page. (It's too long, so I've simplified some of it.)

a = ["70", "80", "90", "100", "110"]
b = [112, 1513, 14, 505, 36]
c = ["url_1", "url_2", "url_3", "url_4", "url_5"]

last = []
num = 0
for l in a:
    if 110 == int(l):
        last.insert(0, c[num])

    elif 100 == int(l):
        last.append(c[num])

    elif 90 == int(l):
        last.append(c[num])

    elif 80 == int(l):
        last.append(c[num])

    elif 70 == int(l):
        last.append(c[num])
    num += 1

print(last)

Originally, it was the idea of putting the elements of variable-a in variable-last in order of magnitude.

But I found that I need to reorder the contents by the elements of list b.

What I want to do is to sort the elements of variable-a in numerical order, while at the same time sorting them again in order of largest in the elements of variable-b.

For example

a = ["70", "100", "90", "100", "100"]
b = [112, 1513, 14, 505, 36]
c = ["url_1", "url_2", "url_3", "url_4", "url_5"]

When classifying '100' in a above, it is also to consider the size of the numbers on the same index in b and classify them in order of size. So, again, the elements of c with the same index are put into the variable last in order. so finally

last = ["url_2", "url_4", "url_5", "url_3", "url_1"]

I want to complete a list in this order. All day long, I failed. help

I suggest you learn about dictionaries and classes.

Code-Apprentice
– Code-Apprentice

2022-08-02 03:46:53 +00:00
Commented Aug 2, 2022 at 3:46 — Code-Apprentice
– Code-Apprentice, Commented Aug 2, 2022 at 3:46

Hampus Larsson · Accepted Answer · 2022-08-01 14:23:36Z

1

You can do this using built-ins:

>>> a = ["70", "100", "90", "100", "100"]
>>> b = [112, 1513, 14, 505, 36]
>>> c = ["url_1", "url_2", "url_3", "url_4", "url_5"]
>>>
>>> sorted(zip(map(int, a), b, c), key=lambda x: x[:2], reverse=True)
[(100, 1513, 'url_2'), (100, 505, 'url_4'), (100, 36, 'url_5'), (90, 14, 'url_3'), (70, 112, 'url_1')]

Then if you want to extract only the "urls":

>>> x = sorted(zip(map(int, a), b, c), key= lambda x: x[:2], reverse=True)
>>> [i[2] for i in x]
['url_2', 'url_4', 'url_5', 'url_3', 'url_1']

The way sorted() works, is that it tries to compare each value, index for index. That means that you can give it a way to sort values based on different "columns" of the sliced value.

You can customize that sorting via the key keyword. You can give it a function (or as I did above, a lambda function).

edited Aug 1, 2022 at 14:23

answered Aug 1, 2022 at 14:18

Hampus Larsson

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1 Comment

anfwkdrn Over a year ago

Thank you. I think this code could solve it.

Anuj Kumar · Accepted Answer · 2022-08-01 14:15:47Z

1

Use pandas dataframe functions:

a = ["70", "100", "90", "100", "100"]
b = [112, 1513, 14, 505, 36]
c = ["url_1", "url_2", "url_3", "url_4", "url_5"]

df = pd.DataFrame([int(i) for i in a], columns=['a'])
df['b'] = b
df['c'] = c

df.sort_values(by = ['a','b'], ascending=False, inplace=True)
print(df['c'].to_list())

answered Aug 1, 2022 at 14:15

Anuj Kumar

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2 Comments

anfwkdrn Over a year ago

Thank you. It's a pity that I can only choose one. I recommend it.

Anuj Kumar Over a year ago

Thanks, good to hear that you found it useful

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