Remove an elements from the string if it contains a stopwords [duplicate]

This question already has an answer here:

How to remove items from a list that contains words found in items in another list [duplicate]

4 answers

I have a list as below:

lst = ['for Sam', 'Just in', 'Mark Rich']

I am trying to remove an element from list of strings(string contains one or more words) which contains stopwords.

As 1st and 2nd elements in the list contains for and in which are stopwords, it will return

new_lst = ['Mark Rich']

What I tried

from nltk.corpus import stopwords



stop_words = set(stopwords.words('english'))



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split(" ") for i in lst]

new_lst = [" ".join(i) for i in new_lst for j in i if j not in stop_words]

Which is giving me output as:

['for Sam', 'Just in', 'Mark Rich', 'Mark Rich']

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

marked as duplicate by Kasrâmvd python
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Jan 2 at 11:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment |

This question already has an answer here:

How to remove items from a list that contains words found in items in another list [duplicate]

4 answers

I have a list as below:

lst = ['for Sam', 'Just in', 'Mark Rich']

I am trying to remove an element from list of strings(string contains one or more words) which contains stopwords.

As 1st and 2nd elements in the list contains for and in which are stopwords, it will return

new_lst = ['Mark Rich']

What I tried

from nltk.corpus import stopwords



stop_words = set(stopwords.words('english'))



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split(" ") for i in lst]

new_lst = [" ".join(i) for i in new_lst for j in i if j not in stop_words]

Which is giving me output as:

['for Sam', 'Just in', 'Mark Rich', 'Mark Rich']

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

marked as duplicate by Kasrâmvd python
Users with the python badge can single-handedly close python questions as duplicates and reopen them as needed.

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Jan 2 at 11:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment |

This question already has an answer here:

How to remove items from a list that contains words found in items in another list [duplicate]

4 answers

I have a list as below:

lst = ['for Sam', 'Just in', 'Mark Rich']

I am trying to remove an element from list of strings(string contains one or more words) which contains stopwords.

As 1st and 2nd elements in the list contains for and in which are stopwords, it will return

new_lst = ['Mark Rich']

What I tried

from nltk.corpus import stopwords



stop_words = set(stopwords.words('english'))



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split(" ") for i in lst]

new_lst = [" ".join(i) for i in new_lst for j in i if j not in stop_words]

Which is giving me output as:

['for Sam', 'Just in', 'Mark Rich', 'Mark Rich']

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

This question already has an answer here:

How to remove items from a list that contains words found in items in another list [duplicate]

4 answers

I have a list as below:

lst = ['for Sam', 'Just in', 'Mark Rich']

I am trying to remove an element from list of strings(string contains one or more words) which contains stopwords.

As 1st and 2nd elements in the list contains for and in which are stopwords, it will return

new_lst = ['Mark Rich']

What I tried

from nltk.corpus import stopwords



stop_words = set(stopwords.words('english'))



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split(" ") for i in lst]

new_lst = [" ".join(i) for i in new_lst for j in i if j not in stop_words]

Which is giving me output as:

['for Sam', 'Just in', 'Mark Rich', 'Mark Rich']

This question already has an answer here:

How to remove items from a list that contains words found in items in another list [duplicate]

4 answers

python python-3.x nltk

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

asked Jan 2 at 11:23

AkshayNevrekar

4,91291840

marked as duplicate by Kasrâmvd python
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Jan 2 at 11:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Kasrâmvd python
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Jan 2 at 11:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment |

2 Answers
2

active

oldest

votes

You need an if statement rather than extra nesting:

new_lst = [' '.join(i) for i in new_lst if not any(j in i for j in stop_words)]

If you wish to utilize set, you can use set.isdisjoint:

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]

Here's a demonstration:

stop_words = {'for', 'in'}



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split() for i in lst]

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]



print(new_lst)



# ['Mark Rich']

edited Jan 2 at 11:34

answered Jan 2 at 11:27

jpp

101k2164115

Your 1st answer worked like a charm but 2nd one is giving an empty list.

– AkshayNevrekar
Jan 2 at 11:33

1

@Sociopath, Nope, works fine, see my example.

– jpp
Jan 2 at 11:34

add a comment |

You can use a list comprehension and use sets to check if any words within the two lists intersect:

[i for i in lst if not set(stop_words) & set(i.split(' '))]

['Mark Rich']]

edited Jan 2 at 11:35

answered Jan 2 at 11:30

yatu

12.9k31341

thanks. Worked like a charm. Only one thing, in your answer you misplaced ]

– AkshayNevrekar
Jan 2 at 11:34

Note set.intersection has higher complexity vs set.disjoint. It's not necessary to calculate the exact intersection of the 2 sets to know if the intersection is empty.

– jpp
Jan 2 at 11:35

1

Yes I actually thought about .isdisjoint right when i saw your answer. thx for clarifying

– yatu
Jan 2 at 11:37

add a comment |

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

You need an if statement rather than extra nesting:

new_lst = [' '.join(i) for i in new_lst if not any(j in i for j in stop_words)]

If you wish to utilize set, you can use set.isdisjoint:

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]

Here's a demonstration:

stop_words = {'for', 'in'}



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split() for i in lst]

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]



print(new_lst)



# ['Mark Rich']

edited Jan 2 at 11:34

answered Jan 2 at 11:27

jpp

101k2164115

Your 1st answer worked like a charm but 2nd one is giving an empty list.

– AkshayNevrekar
Jan 2 at 11:33

1

@Sociopath, Nope, works fine, see my example.

– jpp
Jan 2 at 11:34

add a comment |

You need an if statement rather than extra nesting:

new_lst = [' '.join(i) for i in new_lst if not any(j in i for j in stop_words)]

If you wish to utilize set, you can use set.isdisjoint:

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]

Here's a demonstration:

stop_words = {'for', 'in'}



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split() for i in lst]

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]



print(new_lst)



# ['Mark Rich']

edited Jan 2 at 11:34

answered Jan 2 at 11:27

jpp

101k2164115

Your 1st answer worked like a charm but 2nd one is giving an empty list.

– AkshayNevrekar
Jan 2 at 11:33

1

@Sociopath, Nope, works fine, see my example.

– jpp
Jan 2 at 11:34

add a comment |

You need an if statement rather than extra nesting:

new_lst = [' '.join(i) for i in new_lst if not any(j in i for j in stop_words)]

If you wish to utilize set, you can use set.isdisjoint:

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]

Here's a demonstration:

stop_words = {'for', 'in'}



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split() for i in lst]

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]



print(new_lst)



# ['Mark Rich']

edited Jan 2 at 11:34

answered Jan 2 at 11:27

jpp

101k2164115

You need an if statement rather than extra nesting:

new_lst = [' '.join(i) for i in new_lst if not any(j in i for j in stop_words)]

If you wish to utilize set, you can use set.isdisjoint:

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]

Here's a demonstration:

stop_words = {'for', 'in'}



lst = ['for Sam', 'Just in', 'Mark Rich']

new_lst = [i.split() for i in lst]

new_lst = [' '.join(i) for i in new_lst if stop_words.isdisjoint(i)]



print(new_lst)



# ['Mark Rich']

edited Jan 2 at 11:34

answered Jan 2 at 11:27

jpp

101k2164115

edited Jan 2 at 11:34

answered Jan 2 at 11:27

jpp

101k2164115

answered Jan 2 at 11:27

jpp

101k2164115

answered Jan 2 at 11:27

jpp

101k2164115

Your 1st answer worked like a charm but 2nd one is giving an empty list.

– AkshayNevrekar
Jan 2 at 11:33

1

@Sociopath, Nope, works fine, see my example.

– jpp
Jan 2 at 11:34

add a comment |

Your 1st answer worked like a charm but 2nd one is giving an empty list.

– AkshayNevrekar
Jan 2 at 11:33

1

@Sociopath, Nope, works fine, see my example.

– jpp
Jan 2 at 11:34

Your 1st answer worked like a charm but 2nd one is giving an empty list.

– AkshayNevrekar
Jan 2 at 11:33

@Sociopath, Nope, works fine, see my example.

– jpp
Jan 2 at 11:34

add a comment |

You can use a list comprehension and use sets to check if any words within the two lists intersect:

[i for i in lst if not set(stop_words) & set(i.split(' '))]

['Mark Rich']]

edited Jan 2 at 11:35

answered Jan 2 at 11:30

yatu

12.9k31341

thanks. Worked like a charm. Only one thing, in your answer you misplaced ]

– AkshayNevrekar
Jan 2 at 11:34

Note set.intersection has higher complexity vs set.disjoint. It's not necessary to calculate the exact intersection of the 2 sets to know if the intersection is empty.

– jpp
Jan 2 at 11:35

1

Yes I actually thought about .isdisjoint right when i saw your answer. thx for clarifying

– yatu
Jan 2 at 11:37

add a comment |

You can use a list comprehension and use sets to check if any words within the two lists intersect:

[i for i in lst if not set(stop_words) & set(i.split(' '))]

['Mark Rich']]

edited Jan 2 at 11:35

answered Jan 2 at 11:30

yatu

12.9k31341

thanks. Worked like a charm. Only one thing, in your answer you misplaced ]

– AkshayNevrekar
Jan 2 at 11:34

Note set.intersection has higher complexity vs set.disjoint. It's not necessary to calculate the exact intersection of the 2 sets to know if the intersection is empty.

– jpp
Jan 2 at 11:35

1

Yes I actually thought about .isdisjoint right when i saw your answer. thx for clarifying

– yatu
Jan 2 at 11:37

add a comment |

You can use a list comprehension and use sets to check if any words within the two lists intersect:

[i for i in lst if not set(stop_words) & set(i.split(' '))]

['Mark Rich']]

edited Jan 2 at 11:35

answered Jan 2 at 11:30

yatu

12.9k31341

You can use a list comprehension and use sets to check if any words within the two lists intersect:

[i for i in lst if not set(stop_words) & set(i.split(' '))]

['Mark Rich']]

edited Jan 2 at 11:35

answered Jan 2 at 11:30

yatu

12.9k31341

edited Jan 2 at 11:35

answered Jan 2 at 11:30

yatu

12.9k31341

answered Jan 2 at 11:30

yatu

12.9k31341

answered Jan 2 at 11:30

yatu

12.9k31341

thanks. Worked like a charm. Only one thing, in your answer you misplaced ]

– AkshayNevrekar
Jan 2 at 11:34

Note set.intersection has higher complexity vs set.disjoint. It's not necessary to calculate the exact intersection of the 2 sets to know if the intersection is empty.

– jpp
Jan 2 at 11:35

1

Yes I actually thought about .isdisjoint right when i saw your answer. thx for clarifying

– yatu
Jan 2 at 11:37

add a comment |

thanks. Worked like a charm. Only one thing, in your answer you misplaced ]

– AkshayNevrekar
Jan 2 at 11:34

Note set.intersection has higher complexity vs set.disjoint. It's not necessary to calculate the exact intersection of the 2 sets to know if the intersection is empty.

– jpp
Jan 2 at 11:35

1

Yes I actually thought about .isdisjoint right when i saw your answer. thx for clarifying

– yatu
Jan 2 at 11:37

thanks. Worked like a charm. Only one thing, in your answer you misplaced ]

– AkshayNevrekar
Jan 2 at 11:34

Note set.intersection has higher complexity vs set.disjoint. It's not necessary to calculate the exact intersection of the 2 sets to know if the intersection is empty.

– jpp
Jan 2 at 11:35

Yes I actually thought about .isdisjoint right when i saw your answer. thx for clarifying

– yatu
Jan 2 at 11:37

add a comment |

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