Terry--gmail wrote:
> Marc, my understanding is, is that:
>
> lens[col].append(len(item))
>
> -should be building a mirror image of my list of lists called catalog2,
> which currently has 9 columns by x number of rows, and that we are
> plugging into these positions, the sizes of all the elements in that
> block of data.
>
> If that is true, then lens[col] is creating lists which contain the
> sizes of the elements in that particular column.
>
> So, I completely don't understand why we would eliminate the positioning
> of which list we are referencing in lens by saying:
>
> lens.append(len(item))
>
> It seems to me that, that statement would put the entire block of
> element sizes into one list, and the next MAX statement would then yield
> only a single number, which would be the largest size element it
> encounted in the whole of catalog2!
>
> Or am I really missing the boat here? :)
[Alan Gauld]
> lens = [] * len(catalog2[0])
> for row in catalog2[1:]: # miss the first row
> for col,item in row:
> lens[col].append(len(item))
>
> lens = [max(col) for col in lens]
In the above snipped Alan introduced introduced another bug that wasn't
fixed in any of the posts I saw: lens should be a list of lists.
With "lens" referring to
[[], [], [], ...]
lens[n] refers to the n-th list and thus
lens[col].append(len(item))
appends the items's length to the respective column/list.
The complete fixed code is then:
# make a list with one empty list per column
lens = []
for item in catalog2[0]:
lens.append([])
for row in catalog2:
for col, item in enumerate(row):
lens[col].append(len(item))
lens = [max(col) for col in lens]
The first for-loop can also be written as a list comprehension:
lens = [[] for item in catalog2[0]]
You might be tempted to simplify further
# WRONG:
lens = [[]] * len(catalog2[0])
but that repeats the same inner list len(catalog2[0]) times where you
actually need len(catalog2[0]) distinct lists.
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