go one step back and one step forward in a loop with python

一世执手 提交于 2019-12-02 07:07:10

I wondered how to manage a list (nonsense) like this:

words = ['Bien', '*', 'venue', 'pour', 'les','engage', '*', 'ment', 'trop', 'de', 'YIELD', 'peut','être','contre', '*', 'productif' ]

So I came u with a method like this:

def join_asterisk(ary):
  i, size = 0, len(ary)
  while i < size-2:
    if ary[i+1] == '*':
      yield ary[i] + ary[i+2]
      i+=2
    else: yield ary[i]
    i += 1
  if i < size:
    yield ary[i]

Which returns:

print(list(join_asterisk(words)))
#=> ['Bienvenue', 'pour', 'les', 'engagement', 'trop', 'de', 'YIELD', 'peut', 'être', 'contreproductif']

Instead of thinking "time travel" (i.e. go back and forth), the Pythonic way would be to think functional (time travel has it's place in very resource constrained environments).

One way is to go the enumeration way as @Yosufsn showed. Another is to zip the list with itself, but with padding appended on either side. Like this:

words = ['les','engage', '*', 'ment', 'de','la'] 
for a,b,c in zip([None]*2+words, [None]+words+[None], words+[None]*2):
    if b == '*':
        print( a+c )

I think you need a simple code like this:

words = ['les','engage', '*', 'ment', 'de','la']

for n,word in enumerate (words):
    if word == "*":
        exp = words[n-1] + words[n+1]
        print (exp)

Output:

"engagement"

With this output, you can subsequently check with your dictionary.

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