Python iteration problem!

Question

I have this code here, it should remove the common letters from lists n1 and n2. But when I run this code, it runs only once, as in it it removes only "a" from n1 and n2 and does not delete "k".

Just to clarify this code, it should always only work for 2 words.

name1 = "abdjek" name2 = "doarhsnk" n1l = list(name1) n2l = list(name2) for i in range(len(n1l)): for j in range(len(n2l)): if n1l[i] == n2l[j]: n1l.pop(i) n2l.pop(j) n1l.append('0') n2l.append('1') 

Ok, wait, it seems to work for the above names, but when I have name1 = "naveen" and name2 = "darshana", it does not work!

Answer 1

A more pythonic approach would be to use set and list understanding.

 name1 = "naveen"; name2 = "darshana" name1_set=set(name1) name2_set=set(name2) clean1=[x for x in name1 if x not in name2_set] clean2=[x for x in name2 if x not in name1_set] clean1.extend(['0']*(len(name1)-len(clean1))) clean2.extend(['1']*(len(name2)-len(clean2))) print clean1,clean2 

set gives us O (1) searches, thereby making the whole process faster, making it O (N) instead of O (N ^ 2).

EDIT: In the light of your later comment that the number of events matters, this is an updated version that takes this into account.

 name1 = "naveen"; name2 = "darshana" count1={} count2={} for x in name1: count1[x]=count1.get(x,0)+1 for x in name2: count2[x]=count2.get(x,0)+1 def remove_dups(name,count,null): clean=[] for x in name: if count.get(x,0): count[x]-=1 else: clean.append(x) clean.extend([null]*(len(name)-len(clean))) return clean clean1=remove_dups(name1,count2,'0') clean2=remove_dups(name2,count1,'1') print clean1,clean2 

It uses a dict to count occurrences. Whenever a character is deleted, the corresponding counter for another name is reduced. The complexity is still O (N).

It prints ['v', 'e', 'e', 'n', '0', '0'] and ['d', 'r', 's', 'h', 'a', 'a', '1', '1'] . Is that what you wanted?

Answer 2

I suggest a much simpler approach:

 def removecommon(name1, name2): common = set(name1).intersection(name2) res1 = ''.join(n for n in name1 if n not in common) res2 = ''.join(n for n in name2 if n not in common) return res1, res2 n1, n2 = removecommon('naveen', 'darshana') print n1, n2 

emits vee drsh if desired.

Edit : as indicated in the OP (in the comment - pls do not forget to edit your question too, about OP!) That he really wants to remove only the first in each word of each common letter, the necessary algorithm is, of course, completely different. A simple approach (possibly if the word length is not too long):

 def removefirstcommon(name1, name2): common = set(name1).intersection(name2) n1 = list(name1) for c in common: n1.remove(c) n2 = list(name2) for c in common: n2.remove(c) return ''.join(n1), ''.join(n2) 

A more complex approach (albeit slower for words of normal length) will be faster for extremely long words (since the next is O (N) and the first is O (square N)):

 def removefirstcommonlongwords(name1, name2): common = set(name1).intersection(name2) def mustrem(c, copycom): res = c not in copycom copycom.discard(c) return res cop = set(common) n1 = [c for c in name1 if mustrem(c, cop)] n2 = [c for c in name2 if mustrem(c, common)] return ''.join(n1), ''.join(n2) 

Answer 3

He works here for me. That is, if I add print statements this way:

 name1 = "abdjek" name2 = "doarhsnk" n1l = list(name1) n2l = list(name2) print "Lists before loop:" print n1l print n2l for i in range(len(n1l)): for j in range(len(n2l)): if n1l[i] == n2l[j]: n1l.pop(i) n2l.pop(j) n1l.append('0') n2l.append('1') print "Lists after loop:" print n1l print n2l 

the characters 'a', 'd' and 'k' are deleted:

 > python test.py Lists before loop: ['a', 'b', 'd', 'j', 'e', 'k'] ['d', 'o', 'a', 'r', 'h', 's', 'n', 'k'] Lists after loop: ['b', 'j', 'e', '0', '0', '0'] ['o', 'r', 'h', 's', 'n', '1', '1', '1'] 

Answer 4

Your code may not work as you expect, as it removes matching letters. For example, you see a, then you delete both of your words ...

Answer 5

Your code most likely does not work, because you publish common letters from anywhere in the list, but add replacements ("0" and "1") to the end of the list. They should be in position i and j respectively.

So the loop should look something like this:

 for i in range(len(n1l)): for j in range(len(n2l)): if n1l[i] == n2l[j] and n1l[i] not in ("0", "1"): print "common letter ", n1l[i] # Replace i-th and j-th element n1l[i] = "0" n2l[j] = "1" 

In any case, there are still “pythonic” methods that are already shown in other answers.

EDIT: Tested and works also with name1 = "naveen" / name2 = "darshana" .

Answer 6

Here is some pretty (IMHO pretty elegant) code that works in O (n). If word 1 has N occurrences of the letter x, it removes the first N x from word 2 (and vice versa) - I think this is what you want, but I could be wrong.

 from collections import defaultdict def build(s, chars_s, chars_t): """Return characters of s, with duplicate characters from t removed.""" for i, char in enumerate(s): indexes_s, indexes_t = chars_s[char], chars_t[char] if len(indexes_s) > len(indexes_t) and i >= indexes_s[len(indexes_t)]: yield char def rm_dup(a, b): """Pairwise remove duplicate letters in a and b.""" chars_a, chars_b = defaultdict(list), defaultdict(list) for i, char in enumerate(a): chars_a[char].append(i) for i, char in enumerate(b): chars_b[char].append(i) return (''.join(build(a, chars_a, chars_b)), ''.join(build(b, chars_b, chars_a))) print rm_dup('naveen', 'darshana')