一种删除列表中仅一个元素的连续重复项的pythonic方法

Question

I know there are many other similar questions posted, but there is a difference in mine that makes it unsolvable with their answers.

I have several lists of characters that may have multiple consecutive spaces, of which I need to keep only one. Repetitions of any other character should remain. I did it in the following way:

myList = ['o', 'e', 'i', ' ', ' ', ' ', 'l', 'k', ' ', ' ', ' ', ' ', ' ', 'j', 'u']
myList_copy = [myList[0]]

for i in range(1, len(myList):
    if not(myList[i] == ' ' and myList[i-1] == ' '):
        myList_copy.append(myList[i])

which successfully gives me

['o', 'e', 'i', ' ', 'l', 'k', ' ', 'j', 'u', ' ']

I don't really think this is a very good, fast way to do it.

I have seen posts like this one (and others) which have similar questions. However, see that I actually need to remove only repeated spaces. Maybe what I need help with is using groupby to do this, but that's why the new post.

Thanks in advance.

Answer 1

Yes,Using groupby is a good idea:

import itertools

myList = ['o', 'e', 'i', ' ', ' ', ' ', 'l', 'k', ' ', ' ', ' ', ' ', ' ', 'j', 'u']
result = [key for key,group in itertools.groupby(myList)])

# ['o', 'e', 'i', ' ', 'l', 'k', ' ', 'j', 'u']

If you want to get another elements also duplicate,you can use this:

myList = ['o', 'e', 'i', 'i' , ' ', ' ', ' ', 'l', 'k', ' ', ' ', ' ', ' ', ' ', 'j', 'u']
result = []
for key,group in itertools.groupby(myList):
    if key != ' ': # ' 'string
        for j in group:
            result.append(j)
    else: result.append(key)
print(result)

Answer 2

Another simple? way to do it:

1. Join each item in the myList to create a string
2. Split the string by whitespace
3. Join with a space
4. Convert the string into a list

myList = ['o', 'e', 'i', ' ', ' ', ' ', 'l', 'k', ' ', ' ', ' ', ' ', ' ', 'j', 'u']

new = list(' '.join(''.join(myList).split()))
print(new)

['o', 'e', 'i', ' ', 'l', 'k', ' ', 'j', 'u']

Answer 3

this is the same as yours but in one line

myList_copy = [myList[x] for x in range(len(myList)) if not(myList[x] == ' ' and myList[x-1] == ' ')]

Answer 4

How about using numpy? Try this code.

import numpy as np
myList = ['o', 'e', 'i', ' ', ' ', ' ', 'l', 'k', ' ', ' ', ' ', ' ', ' ', 'j', 'u']
myList = np.array(myList)
myList = [myList[0]] + list(myList[1:][~((myList[1:] == myList[:-1]) & (myList[1:] == ' '))])
print(myList)

Answer 5

You can use zip in a list comprehension to compare each character with the previous one and exclude spaces that are preceded by another space:

myList = [ c for p,c in zip([""]+myList,myList) if (p,c) != (' ',' ') ]

same approach can be used on a string

myList = [ c for p,c in zip("."+myString, myString) if (p,c) != (' ',' ') ]

but split() would probably be more concise if you have a string and want a string as output:

myString = " ".join(myString.split())

Answer 6

What about using a pandas Series and shifting the results?

import pandas as pd
serie = pd.Series(['o', 'e', 'i', ' ', ' ', ' ', 'l', 'k', ' ', ' ', ' ', ' ', ' ', 'j', 'u'])
index = ~(serie == serie.shift(1))
serie = serie[index]