create a dictionary showing connectivity between items of list python-CodePudding

I have a list as:

['Title', 'Text', 'Title', 'Title', 'Text', 'Title', 'Text', 'List', 'Text', 'Title', 'Text', 'Text']

I want every element to be connected to element 'Title" before the element. For example, Text at index 1 is connected to Title at index 0, Title at index 2 would not be connected to any element, because it has another title after it. Text at index 4 is connected to title 3, similarly Text at position 10,11 will be connected to Title at index 9.

This is the expected output:

{1:0,4:3,6:5,7:5,8:5,10:9,11:9}

How can I do that?

CodePudding user response：

You can use a loop:

l = ['Title', 'Text', 'Title', 'Title', 'Text', 'Title', 'Text', 'List', 'Text', 'Title', 'Text', 'Text']

last = -1
out = {}
for i, v in enumerate(l):
    if v == 'Title':
        last = i
    else:
        out[i] = last
print(out)

Output: {1: 0, 4: 3, 6: 5, 7: 5, 8: 5, 10: 9, 11: 9}

CodePudding user response：

Don't take this as good practice, but I was trying to have fun with the := walrus operator and do it all in one comprehension.

prev = -1
li = ['Title', 'Text', 'Title', 'Title', 'Text', 'Title', 'Text', 'List', 'Text', 'Title', 'Text', 'Text']
li2 = {ix:prev
    for ix, v 
    in enumerate(li) 

    if

    #this first part is always True because of the  77
    # why 77 rather than say 1? to avoid -1 1 => 0 if first is Text
    #but it only stores prev on Title values using the walrus `:=`
    ((prev:= (ix if v == "Title" else prev)) 77) 

    #we only want the Text Values
    and v == "Text" 
}

print(f"{li2=}")

#output:

li2={1: 0, 4: 3, 6: 5, 8: 5, 10: 9, 11: 9}

In some defense of this unintuitive code list comprehensions are usually faster than equivalent for loop constructs. Usually => not in this case, where mozway's answer was faster.

CodePudding user response：

Basic logic is to find all postions of root Title and just place the last smaller than current position to current none root element.

Code:

x=  ['Title', 'Text', 'Title', 'Title', 'Text', 'Title', 'Text', 'List', 'Text', 'Title', 'Text', 'Text']

{i:[j for j, v in enumerate(x[:i]) if v==x[0]][-1] for i,v in enumerate(x) if v!=x[0]}

Output:

{1: 0, 4: 3, 6: 5, 7: 5, 8: 5, 10: 9, 11: 9}

CodePudding user response：

One way you can do this (not very efficient but I do not have much time at them moment):

x = ['Title', 'Text', 'Title', 'Title', 'Text', 'Title', 'Text', 'List', 'Text', 'Title', 'Text', 'Text']
d = {}
for i, ele in enumerate(x):
    if ele != 'Title':
        d[i] = i - x[:i 1][::-1].index('Title')

>>> print(d)
>>> {1: 0, 4: 3, 6: 5, 7: 5, 8: 5, 10: 9, 11: 9}

#oneliner:
{i: i-x[:i 1][::-1].index('Title') for i, ele in enumerate(x) if ele != 'Title'}

But this does not check if there is a 'Title' in your list, i.e. the list should always start with 'Title'.