问题
Here is a list containing duplicates:
l1 = ['a', 'b', 'c', 'a', 'a', 'b']
Here is the desired result:
l1 = ['a', 'b', 'c', 'a_1', 'a_2', 'b_1']
How can the duplicates be renamed by appending a count number?
Here is an attempt to achieve this goal; however, is there a more Pythonic way?
for index in range(len(l1)):
counter = 1
list_of_duplicates_for_item = [dup_index for dup_index, item in enumerate(l1) if item == l1[index] and l1.count(l1[index]) > 1]
for dup_index in list_of_duplicates_for_item[1:]:
l1[dup_index] = l1[dup_index] + '_' + str(counter)
counter = counter + 1
回答1:
In Python, generating a new list is usually much easier than changing an existing list. We have generators to do this efficiently. A dict can keep count of occurrences.
l = ['a', 'b', 'c', 'a', 'a', 'b']
def rename_duplicates( old ):
seen = {}
for x in old:
if x in seen:
seen[x] += 1
yield "%s_%d" % (x, seen[x])
else:
seen[x] = 0
yield x
print list(rename_duplicates(l))
回答2:
I would do something like this:
a1 = ['a', 'b', 'c', 'a', 'a', 'b']
a2 = []
d = {}
for i in a1:
d.setdefault(i, -1)
d[i] += 1
if d[i] >= 1:
a2.append('%s_%d' % (i, d[i]))
else:
a2.append(i)
print a2
回答3:
I think the output you're asking for is messy itself, and so there is no clean way of creating it.
How do you intend to use this new list? Would a dictionary of counts like the following work instead?
{'a':3, 'b':2, 'c':1}
If so, I would recommend:
from collections import defaultdict
d = defaultdict(int) # values default to 0
for key in l1:
d[key] += 1
回答4:
Based on your comment to @mathmike, if your ultimate goal is to create a dictionary from a list with duplicate keys, I would use a defaultdict
from the `collections Lib.
>>> from collections import defaultdict
>>> multidict = defaultdict(list)
>>> multidict['a'].append(1)
>>> multidict['b'].append(2)
>>> multidict['a'].append(11)
>>> multidict
defaultdict(<type 'list'>, {'a': [1, 11], 'b': [2]})
来源:https://stackoverflow.com/questions/2837409/how-to-append-count-numbers-to-duplicates-in-a-list-in-python