查找Python中的转换

给出字符串列表，如何获取字母变化的索引。例如，对于列表letters = ['A','A','A','A','A','A','A','A','A','A','A','A','B','C','C','D','D','B','B','A','A','A','A']，我们需要找到每个字母变化的开始并生成结果列表，不包括字母'X'。

期望的结果是：

[(11, 'A'), (12, 'B'), (13, 'C'), (16, 'D'), (18, 'B'), (20, 'A')]

2. 解决方案

为了解决这个问题，我们提供了几种方法：

方法1：

import itertools
import functools
import operator

letters = ['A','A','A','A','A','A','A','A','A','A','A','A','B','C','C','X','D','X','B','B','A','A','A','A']

groupbysecond = functools.partial(itertools.groupby,key=operator.itemgetter(1))

def transitions(letters):
    #segregate transition and non-transition indices
    grouped = groupbysecond(enumerate(zip(letters,letters[1:])))
    # extract first such entry from each group
    firsts = (next(l) for k,l in grouped)
    # group those entries together - where multiple, there are first and last
    # indices of the run of letters
    regrouped = groupbysecond((n,a) for n,(a,b) in firsts)
    # special case for first entry, which wants last index of first letter
    kfirst,lfirst = next(regrouped)
    firstitem = (tuple(lfirst)[-1],) if kfirst != 'X' else ()
    #return first item, and first index for all other letters
    return itertools.chain(firstitem,(next(l) for k,l in regrouped if k != 'X'))

方法2：

firstChangeFound = False
result = []

for i in range(len(letters)):
    if letters[i]!='X' and letters[i]!=letters[i-1]:
        if not firstChangeFound:
            result.append((i-1, letters[i-1])) #Grab the last occurrence of the first character
            result.append((i, letters[i]))
            firstChangeFound = True
        else:
             result.append((i, letters[i])) 

方法3：

for i, letter in enumerate(letters[1:], 1):
    if letter != 'X' and letters[i] != letters[i-1]:
        result.append((i, letter))

first_change = result[0][0]
first_stretch = ''.join(letters[:first_change]).rstrip('X')
if first_stretch:
    result.insert(0, (len(first_stretch) - 1, first_stretch[-1]))

方法4：

def find_transitions(letters):
  transitions = []

  # Iterate over the letters, starting from the second letter.
  for i in range(1, len(letters)):
    # If the current letter is different from the previous letter and not 'X',
    # it's a transition.
    if letters[i] != letters[i - 1] and letters[i] != 'X':
      # Append the index and the letter to the transitions list.
      transitions.append((i, letters[i]))

  # If there are any transitions, return the list of transitions.
  if transitions:
    return transitions

  # Otherwise, return an empty list.
  else:
    return []

方法5：

def find_transitions(letters):
  # Create a list to store the transitions.
  transitions = []

  # Keep track of the current letter and its index.
  current_letter = letters[0]
  current_index = 0

  # Iterate over the letters, starting from the second letter.
  for i in range(1, len(letters)):
    # If the current letter is different from the previous letter and not 'X',
    # it's a transition.
    if letters[i] != current_letter and letters[i] != 'X':
      # Append the index and the letter to the transitions list.
      transitions.append((current_index, current_letter))

      # Update the current letter and its index.
      current_letter = letters[i]
      current_index = i

  # Append the index and the letter for the last transition.
  transitions.append((current_index, current_letter))

  # Return the list of transitions.
  return transitions

这些方法都能够找到所需的转换，你可以根据自己的需要选择最合适的一种。