Srinivas Iyyer wrote:
> Dear group:
>
> I have two lists and I have to match the element in
> the first list to element in second list and print the
> line from second list:
>
> Example:
> listA =['apple','baby','cat']
> listB =['fruit\tapple','tree\tapple','fruit\tmango',
> 'human\tbaby'
> 'infant\tbaby'
> 'human\tAlbert'
> 'animal\tcat'
> 'tiger\tcat'
> 'animan\tdog']
>
>
> I have to take apple, search in listB and print
>
> fruit apple
> tree apple
> infant baby
> human baby
> animal cat
> tiger cat
Presumably that is the results from searching for 'apple', 'baby' and 'cat'...
I would make a helper dictionary that maps from the second element in listB to
a list of listB entries containing that element. This way you make just one
pass over listB:
listA =['apple','baby','cat']
listB =['fruit\tapple','tree\tapple','fruit\tmango',
'human\tbaby',
'infant\tbaby',
'human\tAlbert',
'animal\tcat',
'tiger\tcat',
'animan\tdog']
# Make a helper dict that maps the second element of listB to a list of elements
d={}
for m in listB:
cols = m.split('\t')
term = cols[1]
d.setdefault(term, []).append(m)
for i in listA:
items = d.get(i)
for item in items:
print item
The only tricky part is the line
d.setdefault(term, []).append(m)
This is just a shortcut for something like
try:
data = d[term]
except KeyError:
d[term] = data = []
data.append(m)
Kent
>
>
> I am doing it this way:
>
> for i in listA:
> for m in listB:
> cols = m.split('\t')
> term = cols[2]
> if i == term:
> print m
>
> this is very time consuming.
>
> Because the two columns in listB are redundant I am
> unable to do a dict method.
>
> The other way could be using a reg.ex method (which I
> did not try yet because the script written based on
> equality is still running.
>
> for i in listA:
> pat = re.compile(i)
> for m in listB:
> cols = m.split('\t')
> terms = cols[1]
> if pat.match(terms):
> print m
>
> Experts, could you please suggest any other method
> which is fast and does the job correctly.
>
> Thank you.
>
> Srini
>
>
>
>
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