在计算Python时列出索引超出范围

Question

I have a little problem - I need to iterate through a list of lists (lines in a file) and count how many times the lines begin with six or seven. It was no problem, but I also need to remove the items which repeat themselves, eg if I have this: [6,6,7,7,6], I need to do this out of it: [6,7,6] - it is the number of switches that counts. But somehow the list index is always out of range. Why? Thank you!

def number_of_switches(list):
    counter = []
    for element in li: 
        if int(element[0]) == 6 or int(element[0]) == 7: counter.append(element[0])
        else: pass
    i = 0   
    for i in (0, len(counter)):
    if counter[i] == counter[i+1]: counter.remove(counter[i+1])
    print 'number of switches is', len(counter)-1 #the first number doesn't count as switch, hence '-1'

Answer 1

for i in (0, len(counter)) only iterates over two elements: 0, and the length of counter . It does not count up from 0 to the length. For that you need range:

for i in range(len(counter))

However, you should not do that either. Think about what happens each time you remove an element: the list is now one shorter, but you are iterating until the end of the original list. So you will quickly get an IndexError. You need to append the non-matching elements to a new list, rather than removing from the current one.

Answer 2

You should be doing for i in range(len(counter)): and also if counter[i] == counter[i+1]: will give an error in this range. You need to handle the end case.

This might be a better solution:

def number_of_switches(list):
    counter = []
    prev = li[0]
    for element in range(1, len(li)): 
        if int(li[element][0]) != prev:
            count.append(int(li[element][0]))
    print 'number of switches is', len(counter)-1 #the first number doesn't count as switch, hence '-1'

Answer 3

When you write:

for i in (0, len(counter)):

That means for in those two values: 0 and len(counter) and, len(counter) is the index out of range. I think you meant:

for i in range(0, len(counter)):