如何在python中重命名字符串重复项？

Question

I am trying to replace duplicate string with duplicate integer. For example:

mylist = ["name", "state", "name", "city", "city", "name", "zip", "zip", "name"]

I want the list as

mylist = [1, 'State', 1, 2,2,1,3,3,1]

All name renamed as 1, city as 2 and zip 3. If there are more duplicates those will be also renamed as 4, 5 6 etc.

I have tried with

mylist = ["name", "state", "name", "city", "name", "zip", "zip"]
from collections import Counter 
counts = Counter(mylist) 
for s,num in counts.items():
    if num > 1:
       mylist[mylist.index(s)] = 1

But got

mylist = [1, 'state', 'name', 'city', 'name', 1, 'zip']

How to get 1 for name, 2 for city, 3 for zip and 4 for next duplicate value?

Answer 1

Just modified your code

mylist = ["name", "state", "name", "city", "city", "name", "zip", "zip", "name"]
from collections import Counter
counts = Counter(mylist)
counts
Out[309]: Counter({'city': 1, 'name': 3, 'state': 1, 'zip': 2})
Count=1
for s,num in counts.items():
    if num > 1:
       for  i, j in enumerate(mylist):
           if j==s:
               mylist[i] = Count
       Count=Count+1
mylist
Out[320]: [1, 'state', 1, 2, 2, 1, 3, 3, 1]

Answer 2

Almost there! I commented the additional code:

from collections import Counter 

mylist = ["name", "state", "name", "city", "city", "name", "zip", "zip", "name"]

counts = Counter(mylist) 

c = 0

for s,num in counts.items():
    if num > 1:
      c+= 1 # create a variable (integer) to replace the var in the list (starting with 1 as in your example)
      for x in mylist: # since index returns only the first instance, iterate over your list
        if x == s: 
          mylist[mylist.index(x)] = c # replace with your new integer variable

print(mylist)
# [1, 'state', 1, 2, 2, 1, 3, 3, 1]

Answer 3

Construct a dictionary of indices for each item in mylist - this is similar to using collections.Counter except it preserves the item indices. Use an OrderedDict to preserve the order of the items in the list.

import collections
mylist = ["name", "state", "name", "city", "city", "name", "zip", "zip", "name"]       
d = collections.OrderedDict()
for index, item in enumerate(mylist):
    try:
        d[item].append(index)
    except KeyError:
        d[item] = [index]

Iterate over the dictionary values; check the length; change the item if the criteria is met.

count = 1
for indices in d.values():
    if len(indices) > 1:
        for index in indices:
            mylist[index] = count
        count+=1
print(mylist)

Answer 4

Maybe not the most beautiful solution, but this works:

mylist = ["name", "state", "name", "city", "city", "name", "zip", "zip"]
from collections import Counter 
counts = Counter(mylist) 
val = 1
for s,num in counts.items():
    if num > 1:
        counts[s] = val
        val += 1
    else:
        counts[s] = 0
mylist = [x if counts[x]==0 else counts[x] for x in mylist]
mylist

You get then [1, 'state', 1, 2, 2, 1, 3, 3]

Answer 5

You're very close...

for s in counts:
    if counts[s] > 1: 
        mylist[mylist.index(s)] = mylist.index(s)
# myList is now [0, 'state', 2, 'city', 'name', 5, 6]

Answer 6

You can use this solution:

from collections import Counter
from itertools import count
from operator import itemgetter

mylist = ["name", "state", "name", "city", "city", "name", "zip", "zip", "name"]
C = Counter(mylist)
c = count(start=1)
C = {k: next(c) if v > 1 else k for k, v in C.items()}
itemgetter(*mylist)(C)
# (1, 'state', 1, 2, 2, 1, 3, 3, 1)