On 7 Dez 2004, [EMAIL PROTECTED] wrote:
> I have two lists names x and seq.
>
> I am trying to find element of x in element of seq. I
> find them. However, I want to print element in seq
> that contains element of x and also the next element
> in seq.
[...]
> 3. TRIAL 3:
> I just asked to print the element in seq that matched
> element 1 in X. It prints only that element, however
> I want to print the next element too and I cannot get
> it.
>>>> for ele1 in x:
> for ele2 in seq:
> if ele1 in ele2:
> print ele2
>
[...]
>>>> len(x)
> 4504
>>>> x[1:10]
> ['454:494', '319:607', '319:608', '322:289',
> '322:290', '183:330', '183:329', '364:95', '364:96']
>>>> len(seq)
> 398169
>>>> seq[0:4]
> ['>probe:HG-U95Av2:1000_at:399:559;
> Interrogation_Position=1367; Antisense;',
> 'TCTCCTTTGCTGAGGCCTCCAGCTT',
> '>probe:HG-U95Av2:1000_at:544:185;
> Interrogation_Position=1379; Antisense;',
> 'AGGCCTCCAGCTTCAGGCAGGCCAA']
[...]
> How Do I WANT:
>
> I want to print get an output like this:
>
>
>>probe:HG-U95Av2:1000_at:399:559;
> Interrogation_Position=1367; Antisense;'
> TCTCCTTTGCTGAGGCCTCCAGCTT
>
>>probe:HG-U95Av2:1000_at:544:185;
> Interrogation_Position=1379; Antisense;
> AGGCCTCCAGCTTCAGGCAGGCCAA
Hi, you got some replies how to do it, but IMO there are two other
possibilities:
(a) Turn seq into a dictionary with the parts of the string that are
matched against from list x as keys. Since seq is long that may be
much faster.
def list_to_dict (lst):
d = {}
reg = re.compile(':.+?:.+?:(.+?:.+?);')
for val1, val2 in lst:
key = reg.search(val1).group(1)
d[key] = val1 + val2
return d
import re
seq_dic = list_to_dict(zip(seq[::2], seq[1::2]))
for key in x:
val = seq_dic.get(key)
if val: print val
The above function uses a regular expression to extract the part of
the string you are interested in and uses it as key in a dictionary.
To find the corrresponding list entries `zip(seq[::2], seq[1::2])'
is used; seq[::2] is the first, the third, the fifth ... entry of
the list and seq[1::2] is the second, the fourth, the sixth entry of
the list. zip() packs them together in a tuple.
(b) If you care about memory iterate about seq with izip (from
itertools).
from itertools import izip as izip
reg = re.compile(':.+?:.+?:(.+?:.+?);')
for val1, val2 in izip(seq[::2], seq[1::2]):
if reg.search(val1).group(1) in x:
print val1, val2
Instead of zip() izip() is here used (it does not create the whole
list at once). Alöso no dictionary is used. What's better for you
shows only testing.
Karl
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