Why this code is not working properly for keyword search?

I have read the following code from kaggle exercises. The aim of the code for the function multi_word_search(documents1, keywords1) is to show the string indices in documents1 that contain certain words in keywords1. For example if

documents1 = ['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords1 = ['want', 'your']

then the output of the function should be {'want': [0, 2], 'your': [1]} but unfortunately the output that I get after running the code is {'want': [0, 2]}. Where is the problem in the code. Any help in this regard will be much appreciated. Thanks in advance.

def word_search(documents, keyword):

    indices=

    for i, doc in enumerate(documents):

        tokens=doc.split()

        normalized=[token.rstrip('.,').lower() for token in tokens]

        if keyword.lower() in normalized:

            indices.append(i)

    return indices

def multi_word_search(documents1, keywords1):

    keyword_to_indices={}

    for keyword2 in keywords1:

        keyword_to_indices[keyword2]=word_search(documents1, keyword2)

    return keyword_to_indices

s=['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords=['want', 'your']

r=multi_word_search(s,keywords)

print(r)

edited Nov 12 at 1:39

wwii

10.3k31744

asked Nov 12 at 0:43

Frank Moses

1112

1

Please fix the indentation for your functions.
– wwii
Nov 12 at 1:05

2

I get your expected result, {'want': [0, 2], 'your': [1]}, with the code you posted. In your posted code, keywords contains 'your' - BUT in your explanation of the problem you stated keywords1 contains 'you', which kewords are you using?
– wwii
Nov 12 at 1:14

1

your code works fine. there are improvements you can make, but you will get your expected output
– aydow
Nov 12 at 1:25

1

If you were trying to use libraries without importing them it would be getting NameErrors
– wwii
Nov 12 at 1:25

1

If you are curious, you can watch your code execute at pythontutor.com
– wwii
Nov 12 at 1:26

|
show 8 more comments

documents1 = ['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords1 = ['want', 'your']

def word_search(documents, keyword):

    indices=

    for i, doc in enumerate(documents):

        tokens=doc.split()

        normalized=[token.rstrip('.,').lower() for token in tokens]

        if keyword.lower() in normalized:

            indices.append(i)

    return indices

def multi_word_search(documents1, keywords1):

    keyword_to_indices={}

    for keyword2 in keywords1:

        keyword_to_indices[keyword2]=word_search(documents1, keyword2)

    return keyword_to_indices

s=['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords=['want', 'your']

r=multi_word_search(s,keywords)

print(r)

edited Nov 12 at 1:39

wwii

10.3k31744

asked Nov 12 at 0:43

Frank Moses

1112

1

Please fix the indentation for your functions.
– wwii
Nov 12 at 1:05

2

I get your expected result, {'want': [0, 2], 'your': [1]}, with the code you posted. In your posted code, keywords contains 'your' - BUT in your explanation of the problem you stated keywords1 contains 'you', which kewords are you using?
– wwii
Nov 12 at 1:14

1

your code works fine. there are improvements you can make, but you will get your expected output
– aydow
Nov 12 at 1:25

1

If you were trying to use libraries without importing them it would be getting NameErrors
– wwii
Nov 12 at 1:25

1

If you are curious, you can watch your code execute at pythontutor.com
– wwii
Nov 12 at 1:26

|
show 8 more comments

documents1 = ['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords1 = ['want', 'your']

def word_search(documents, keyword):

    indices=

    for i, doc in enumerate(documents):

        tokens=doc.split()

        normalized=[token.rstrip('.,').lower() for token in tokens]

        if keyword.lower() in normalized:

            indices.append(i)

    return indices

def multi_word_search(documents1, keywords1):

    keyword_to_indices={}

    for keyword2 in keywords1:

        keyword_to_indices[keyword2]=word_search(documents1, keyword2)

    return keyword_to_indices

s=['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords=['want', 'your']

r=multi_word_search(s,keywords)

print(r)

edited Nov 12 at 1:39

wwii

10.3k31744

asked Nov 12 at 0:43

Frank Moses

1112

documents1 = ['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords1 = ['want', 'your']

def word_search(documents, keyword):

    indices=

    for i, doc in enumerate(documents):

        tokens=doc.split()

        normalized=[token.rstrip('.,').lower() for token in tokens]

        if keyword.lower() in normalized:

            indices.append(i)

    return indices

def multi_word_search(documents1, keywords1):

    keyword_to_indices={}

    for keyword2 in keywords1:

        keyword_to_indices[keyword2]=word_search(documents1, keyword2)

    return keyword_to_indices

s=['what do you want to do', 'what is your research goal', 'what do you want to accomplish in life']

keywords=['want', 'your']

r=multi_word_search(s,keywords)

print(r)

python function keyword-search wordsearch

edited Nov 12 at 1:39

wwii

10.3k31744

asked Nov 12 at 0:43

Frank Moses

1112

edited Nov 12 at 1:39

wwii

10.3k31744

asked Nov 12 at 0:43

Frank Moses

1112

edited Nov 12 at 1:39

wwii

10.3k31744

edited Nov 12 at 1:39

wwii

10.3k31744

edited Nov 12 at 1:39

wwii

10.3k31744

asked Nov 12 at 0:43

Frank Moses

1112

asked Nov 12 at 0:43

Frank Moses

1112

asked Nov 12 at 0:43

Frank Moses

1112

1

Please fix the indentation for your functions.
– wwii
Nov 12 at 1:05

2

I get your expected result, {'want': [0, 2], 'your': [1]}, with the code you posted. In your posted code, keywords contains 'your' - BUT in your explanation of the problem you stated keywords1 contains 'you', which kewords are you using?
– wwii
Nov 12 at 1:14

1

your code works fine. there are improvements you can make, but you will get your expected output
– aydow
Nov 12 at 1:25

1

If you were trying to use libraries without importing them it would be getting NameErrors
– wwii
Nov 12 at 1:25

1

If you are curious, you can watch your code execute at pythontutor.com
– wwii
Nov 12 at 1:26

|
show 8 more comments

1

Please fix the indentation for your functions.
– wwii
Nov 12 at 1:05

2

I get your expected result, {'want': [0, 2], 'your': [1]}, with the code you posted. In your posted code, keywords contains 'your' - BUT in your explanation of the problem you stated keywords1 contains 'you', which kewords are you using?
– wwii
Nov 12 at 1:14

1

your code works fine. there are improvements you can make, but you will get your expected output
– aydow
Nov 12 at 1:25

1

If you were trying to use libraries without importing them it would be getting NameErrors
– wwii
Nov 12 at 1:25

1

If you are curious, you can watch your code execute at pythontutor.com
– wwii
Nov 12 at 1:26

Please fix the indentation for your functions.
– wwii
Nov 12 at 1:05

I get your expected result, {'want': [0, 2], 'your': [1]}, with the code you posted. In your posted code, keywords contains 'your' - BUT in your explanation of the problem you stated keywords1 contains 'you', which kewords are you using?
– wwii
Nov 12 at 1:14

your code works fine. there are improvements you can make, but you will get your expected output
– aydow
Nov 12 at 1:25

If you were trying to use libraries without importing them it would be getting NameErrors
– wwii
Nov 12 at 1:25

If you are curious, you can watch your code execute at pythontutor.com
– wwii
Nov 12 at 1:26

|
show 8 more comments

1 Answer
1

active

oldest

votes

-1

Instead of writing such lengthy function, you can use dictionary comprehension to get the same results.

In [12]: d = {i : [n for n, k in enumerate(documents1) if any([i == j for j in k.split()])

...: ] for i in keywords1}



In [13]: d

Out[13]: {'want': [0, 2], 'your': [1]}

Here, I'm running loop over the keywords for dictionary (d) keys and the list is being populated by running another loop to find the index of the list that contains the desired keywords.

answered Nov 12 at 1:54

mamun

7131612

Does that answer the question?
– wwii
Nov 12 at 5:40

@wwii Why downvote? I said in the beginning that instead of writing such long code that is prone to error if the op is curious enough he can learn the dictionary comprehension. That's how I'm learning programming. If someone else comes with a better idea, if it seems interesting I'll learn it.
– mamun
Nov 12 at 5:47

It still is not an answer to the question and whether a dictionary comprehension is less prone to error than a regular for loop is an opinion.
– wwii
Nov 12 at 5:52

Given the input, you can reach the output several different ways. I showed one way. Better or not certainly opinion based, but seeing all the different options and choosing the best one is certainly better than one way of doing things.
– mamun
Nov 12 at 5:59

Maybe I'm being a tad pedantic but this a question and answer site.
– wwii
Nov 12 at 6:10

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

-1

Instead of writing such lengthy function, you can use dictionary comprehension to get the same results.

In [12]: d = {i : [n for n, k in enumerate(documents1) if any([i == j for j in k.split()])

...: ] for i in keywords1}



In [13]: d

Out[13]: {'want': [0, 2], 'your': [1]}

Here, I'm running loop over the keywords for dictionary (d) keys and the list is being populated by running another loop to find the index of the list that contains the desired keywords.

answered Nov 12 at 1:54

mamun

7131612

Does that answer the question?
– wwii
Nov 12 at 5:40

@wwii Why downvote? I said in the beginning that instead of writing such long code that is prone to error if the op is curious enough he can learn the dictionary comprehension. That's how I'm learning programming. If someone else comes with a better idea, if it seems interesting I'll learn it.
– mamun
Nov 12 at 5:47

It still is not an answer to the question and whether a dictionary comprehension is less prone to error than a regular for loop is an opinion.
– wwii
Nov 12 at 5:52

Given the input, you can reach the output several different ways. I showed one way. Better or not certainly opinion based, but seeing all the different options and choosing the best one is certainly better than one way of doing things.
– mamun
Nov 12 at 5:59

Maybe I'm being a tad pedantic but this a question and answer site.
– wwii
Nov 12 at 6:10

add a comment |

-1

Instead of writing such lengthy function, you can use dictionary comprehension to get the same results.

In [12]: d = {i : [n for n, k in enumerate(documents1) if any([i == j for j in k.split()])

...: ] for i in keywords1}



In [13]: d

Out[13]: {'want': [0, 2], 'your': [1]}

Here, I'm running loop over the keywords for dictionary (d) keys and the list is being populated by running another loop to find the index of the list that contains the desired keywords.

answered Nov 12 at 1:54

mamun

7131612

Does that answer the question?
– wwii
Nov 12 at 5:40

@wwii Why downvote? I said in the beginning that instead of writing such long code that is prone to error if the op is curious enough he can learn the dictionary comprehension. That's how I'm learning programming. If someone else comes with a better idea, if it seems interesting I'll learn it.
– mamun
Nov 12 at 5:47

It still is not an answer to the question and whether a dictionary comprehension is less prone to error than a regular for loop is an opinion.
– wwii
Nov 12 at 5:52

Given the input, you can reach the output several different ways. I showed one way. Better or not certainly opinion based, but seeing all the different options and choosing the best one is certainly better than one way of doing things.
– mamun
Nov 12 at 5:59

Maybe I'm being a tad pedantic but this a question and answer site.
– wwii
Nov 12 at 6:10

add a comment |

-1

Instead of writing such lengthy function, you can use dictionary comprehension to get the same results.

In [12]: d = {i : [n for n, k in enumerate(documents1) if any([i == j for j in k.split()])

...: ] for i in keywords1}



In [13]: d

Out[13]: {'want': [0, 2], 'your': [1]}

Here, I'm running loop over the keywords for dictionary (d) keys and the list is being populated by running another loop to find the index of the list that contains the desired keywords.

answered Nov 12 at 1:54

mamun

7131612

Instead of writing such lengthy function, you can use dictionary comprehension to get the same results.

In [12]: d = {i : [n for n, k in enumerate(documents1) if any([i == j for j in k.split()])

...: ] for i in keywords1}



In [13]: d

Out[13]: {'want': [0, 2], 'your': [1]}

Here, I'm running loop over the keywords for dictionary (d) keys and the list is being populated by running another loop to find the index of the list that contains the desired keywords.

answered Nov 12 at 1:54

mamun

7131612

answered Nov 12 at 1:54

mamun

7131612

answered Nov 12 at 1:54

mamun

7131612

answered Nov 12 at 1:54

mamun

7131612

Does that answer the question?
– wwii
Nov 12 at 5:40

@wwii Why downvote? I said in the beginning that instead of writing such long code that is prone to error if the op is curious enough he can learn the dictionary comprehension. That's how I'm learning programming. If someone else comes with a better idea, if it seems interesting I'll learn it.
– mamun
Nov 12 at 5:47

It still is not an answer to the question and whether a dictionary comprehension is less prone to error than a regular for loop is an opinion.
– wwii
Nov 12 at 5:52

Given the input, you can reach the output several different ways. I showed one way. Better or not certainly opinion based, but seeing all the different options and choosing the best one is certainly better than one way of doing things.
– mamun
Nov 12 at 5:59

Maybe I'm being a tad pedantic but this a question and answer site.
– wwii
Nov 12 at 6:10

add a comment |

Does that answer the question?
– wwii
Nov 12 at 5:40

@wwii Why downvote? I said in the beginning that instead of writing such long code that is prone to error if the op is curious enough he can learn the dictionary comprehension. That's how I'm learning programming. If someone else comes with a better idea, if it seems interesting I'll learn it.
– mamun
Nov 12 at 5:47

It still is not an answer to the question and whether a dictionary comprehension is less prone to error than a regular for loop is an opinion.
– wwii
Nov 12 at 5:52

Given the input, you can reach the output several different ways. I showed one way. Better or not certainly opinion based, but seeing all the different options and choosing the best one is certainly better than one way of doing things.
– mamun
Nov 12 at 5:59

Maybe I'm being a tad pedantic but this a question and answer site.
– wwii
Nov 12 at 6:10

Does that answer the question?
– wwii
Nov 12 at 5:40

@wwii Why downvote? I said in the beginning that instead of writing such long code that is prone to error if the op is curious enough he can learn the dictionary comprehension. That's how I'm learning programming. If someone else comes with a better idea, if it seems interesting I'll learn it.
– mamun
Nov 12 at 5:47

It still is not an answer to the question and whether a dictionary comprehension is less prone to error than a regular for loop is an opinion.
– wwii
Nov 12 at 5:52

Given the input, you can reach the output several different ways. I showed one way. Better or not certainly opinion based, but seeing all the different options and choosing the best one is certainly better than one way of doing things.
– mamun
Nov 12 at 5:59

Maybe I'm being a tad pedantic but this a question and answer site.
– wwii
Nov 12 at 6:10

add a comment |

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