Sorting elements from a list of tuples into two separate lists-CodePudding

As the title says I'm in the process of sorting tagged elements into two lists. My current code for this is:

lst = [("foo", "good"), ("bar", "bad")...("x", "n")]

def sort(items):
    good = []
    bad = []
    for i in range(len(items)):
        if items[i][1] == 'good':
            good  = items[i][0]
        else:
            bad  = items[i][0]
    return good, bad


alpha, beta = sort(lst)

My output is simply f. I've been trying to work this out for a while but the solution has been eluding me. Any suggestions?

Thank you all.

CodePudding user response：

You can fix your code:

def sort(items):
    good = []
    bad = []
    for x, y in items:
        if y == 'good':
            good.append(x)
        else:
            bad.append(x)
    return good, bad

Note that = corresponds to list.extend which adds iterables element-wise. For strings that means char by char.

CodePudding user response：

To create new lists with mapping or filtering, it's often shorter and easier to read to use list comprehensions rather than calling .append in a loop:

data = [("foo", "good"), ("bar", "bad"), ("head", "good"), ("tails", "bad"), ("x", "n")]

good_lst = [x for x,quality in data if quality == 'good']
bad_lst = [x for x,quality in data if quality != 'good']

print(good_lst)
# ['foo', 'head']
print(bad_lst)
# ['bar', 'tails', 'x']

CodePudding user response：

You can try to solve this in a cleaner way using list comprenhension to compute both lists:

def sort(items):
    good = [items[idx][0] for idx in range(0,len(items)) if items[idx][1] == 'good']
    bad = [items[idx][0] for idx in range(0,len(items)) if items[idx][1] == 'bad']

    return good, bad

If performance is not an issue this could be better due to it is a declarative approach instead of an imperative one so it could be easier to read, but takes into account that it iterates the list twice.

Hopes it helps you!

CodePudding user response：

Your code is fine, but you need to change one thing:

items[i][0] -> [items[i][0]]

When you want to use you need to have values as list:

>>> [2,3]   [4]
[2,3,4]

Your code with one change:

lst = [("foo", "good"), ("bar", "bad"), ("head", "good"), ("tails", "bad")]

def sort(items):
    good = []
    bad = []
    for item in items:
        if item[1] == 'good':
            good  = [item[0]]
        else:
            bad  = [item[0]]
    return good, bad


alpha, beta = sort(lst)
print (alpha, beta)

Output:

['foo', 'head'] ['bar', 'tails']

CodePudding user response：

= is used when you try to add LHS to RHS with same datatype. When you do good = items[i][0] it's like adding foo as a list (['f', 'o', 'o']) to the list good. That's the reason you see the behavior. Hence use list.append() to add element to a list. Try out this:

lst = [("foo", "good"), ("bar", "bad"), ("head", "good"), ("tails", "bad")]

def sort(items):
    good = []
    bad = []
    for item in items:
        #print (item, type(item))
        if item[1] == 'good':
            good.append(item[0])
        else:
            bad.append(item[0])
    return good, bad


alpha, beta = sort(lst)
print (alpha, beta)

Output:

['foo', 'head'] ['bar', 'tails']

CodePudding user response：

I just changed good = items[i][0] to good.append(i[0])

lst = [("foo", "good"),("bar", "bad"),("x", "n")]

def sort(items):
    good = []
    bad = []
    for i in items:
        if i[1] == 'good':
            good.append(i[0])
        else:
            bad.append(i[0])
    return good, bad


alpha, beta = sort(lst)
print(alpha)
print(beta)

Output:

['foo']
['bar', 'x']