Exclude the entry of identical data in a row

Question

There is such a code.

name_elem_site = text_element_by_class_name(driver, "texttexttext") currentList = [] currentWord = currentList.append(str(name_elem_site)) if name_elem_site: for word in currentList: if word == currentWord: continue currentWord = word currentList.append(word) log.write("%s\n" % word) log.flush()

Objective: to obtain data by word and write to the list, without recording a row of running data. Those. should be so

Word1, Word2, Word1, Word3, Word2, Word3, Word2, .., WordN

And I get

Word1, Word1, Word1, Word2, Word2, Word2, Word3, .., WordN

Tried to add if currentList[-1] != word

But the result is the same ...

 name_elem_site = text_element_by_class_name(driver, "texttexttext") currentList = [] currentWord = currentList.append(str(name_elem_site)) if name_elem_site: for word in currentList: if word == currentWord: continue currentWord = word if currentList[-1] != word: # последний элемент не равен текущему currentList.append(word) log.write("%s\n" % word) log.flush()

How to proceed? Beginner, I can not guess.

Guess what your text_element_by_class_name () method returns is none other than you.
@Alexey, I wrote, for example, instead of name_elem_site = Word1.
After a while Slovo2 and so on, there can be repetitions, the main thing in the record is to chop off consecutive identical words.
If name_elem_site = 'Word1', then the rest of the code becomes meaningless, since the currentList in this case always consists of 1 element.
If these are two different code fragments from different parts of the program, then it is worth mentioning this somehow.

Alexey Reytsman Alexey Reytsman 345 2 silver marks 7 bronze marks · Accepted Answer · 2016-12-21T11:06:17

Based on the assumption that you somehow get a list, I can suggest using something like this:

 # Допустим у вас как-то получился вот такой список initial_words_list = ['Слово1', 'Слово2', 'Слово1', 'Слово3', 'Слово3', 'Слово1', 'Слово1', 'Слово1', 'Слово1', 'Слово3', 'Слово2', 'Слово2', 'Слово2', 'Слово1', 'Слово1', 'Слово2', 'Слово3', 'Слово3', 'Слово3', 'Слово2', 'Слово2', 'Слово2', 'Слово2', 'Слово1', 'Слово1', 'Слово1', 'Слово2'] result_list = [] # создадим пустой список, в который будем складывать неповторяющиеся слова previous_word = '' # создаем пустую текстовую переменную, в которую будем записывать предыдущее слово for word in initial_words_list: if word != previous_word: # если текущее слово не равно предыдущему result_list.append(word) # записываем его в список-результат previous_word = word print(result_list) # ['Слово1', 'Слово2', 'Слово1', 'Слово3', 'Слово1', 'Слово3', 'Слово2', 'Слово1', 'Слово2', 'Слово3', 'Слово2', 'Слово1', 'Слово2']

it is easier to get SlovoN and write a value to the file, then open the file and if the last line = the new value received, then skip and wait for the next word.
In the code you provide, the list is first reset to zero, and then one element is added to it.

Answer 2 · 2016-12-21T09:36:25

Try the function groupby from the standard library itertools

 from itertools import groupby l = ['Слово1', 'Слово1', 'Слово2', 'Слово2', 'Слово3', 'Слово4', 'Слово4'] new_l = [word for word, group in groupby(l)] print(new_l) # Напечатает: ['Слово1', 'Слово2', 'Слово3', 'Слово4']

PS: Learn standard libraries. They are not just added to the language itself.

# Prints: ['Word1', 'Word2', 'Word1', 'Word2', 'Word3'] This is necessary.
# So it is impossible: ['Word1', 'Word1', 'Word2', 'Word2', 'Word1']
Removing duplicates through groupby can be replaced by the set: print(set(l))
@ gil9red, new_l = set (l) also removes repetitions altogether l = ['Слово1', 'Слово2', 'Слово2', 'Слово1', 'Слово2', 'Слово3', 'Слово1'] returns only {'Слово3', 'Слово2', 'Слово1'} and I need `{'Word1', 'Word2', 'Word1', 'Word2', 'Word3', 'Word1'}`
@ J.Doe and I said this to Alexander :) And how to solve this issue you wrote in the commentary on the issue
The groupby doesn’t just "collapse" all repetitions, but only those that go in a row.

Exclude the entry of identical data in a row

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