Count the number of words in dictionary from a list

Given a list:

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

and a dictionary

dict = {'orange': 4, 'apple':2, 'pears': 1}

if a string from the list already exist in dict update the value else add a new key and its counting.

result:

dict = {'orange' = 5, 'apple':3, 'pears':3, 'banana':1}

I tried:

count = 0

for string on lst:

    if string in dict.keys():

        for num in dict:

            count = count + num

            num = count

I don't know how to continue

edited Nov 20 '18 at 17:41

iElden

680417

asked Nov 20 '18 at 17:37

Comp

456

add a comment |

Given a list:

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

and a dictionary

dict = {'orange': 4, 'apple':2, 'pears': 1}

if a string from the list already exist in dict update the value else add a new key and its counting.

result:

dict = {'orange' = 5, 'apple':3, 'pears':3, 'banana':1}

I tried:

count = 0

for string on lst:

    if string in dict.keys():

        for num in dict:

            count = count + num

            num = count

I don't know how to continue

edited Nov 20 '18 at 17:41

iElden

680417

asked Nov 20 '18 at 17:37

Comp

456

add a comment |

Given a list:

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

and a dictionary

dict = {'orange': 4, 'apple':2, 'pears': 1}

if a string from the list already exist in dict update the value else add a new key and its counting.

result:

dict = {'orange' = 5, 'apple':3, 'pears':3, 'banana':1}

I tried:

count = 0

for string on lst:

    if string in dict.keys():

        for num in dict:

            count = count + num

            num = count

I don't know how to continue

edited Nov 20 '18 at 17:41

iElden

680417

asked Nov 20 '18 at 17:37

Comp

456

Given a list:

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

and a dictionary

dict = {'orange': 4, 'apple':2, 'pears': 1}

if a string from the list already exist in dict update the value else add a new key and its counting.

result:

dict = {'orange' = 5, 'apple':3, 'pears':3, 'banana':1}

I tried:

count = 0

for string on lst:

    if string in dict.keys():

        for num in dict:

            count = count + num

            num = count

I don't know how to continue

python list dictionary

edited Nov 20 '18 at 17:41

iElden

680417

asked Nov 20 '18 at 17:37

Comp

456

edited Nov 20 '18 at 17:41

iElden

680417

asked Nov 20 '18 at 17:37

Comp

456

edited Nov 20 '18 at 17:41

iElden

680417

edited Nov 20 '18 at 17:41

iElden

680417

edited Nov 20 '18 at 17:41

iElden

680417

asked Nov 20 '18 at 17:37

Comp

456

asked Nov 20 '18 at 17:37

Comp

456

asked Nov 20 '18 at 17:37

Comp

456

add a comment |

4 Answers
4

active

oldest

votes

Your answer was almost correct:

for string in lst:

    if string in dict.keys():

        dict[string] += 1

    else:

        dict[string] = 1

This is assuming that a string you haven't seen yet starts with a value of 1, which seems to be the case given your output.

You could also remove the .keys(), as python will automatically check in the keys for the values you are looping on, hence:

for string in lst:

    if string in dict:

        dict[string] += 1

    else:

        dict[string] = 1

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

add a comment |

You can use collections.Counter

from collections import Counter



>>> lst = ['apple', 'orange', 'pears', 'pears', 'banana']

>>> d = {'orange': 4, 'apple':2, 'pears': 1}



>>> count = Counter(d)

>>> count

Counter({'orange': 4, 'apple': 2, 'pears': 1})

>>> count += Counter(lst)

>>> count

Counter({'orange': 5, 'pears': 3, 'apple': 3, 'banana': 1})

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

I know this is off topic, but how did it place pears ahead of apple?

– Sandesh34
Nov 20 '18 at 17:44

I don't think there is a need of Counter(d), because d itself is a dictionary of counts and you can access with d.get().

– Austin
Nov 20 '18 at 17:47

@Austin Yes but then you still have to write the update logic yourself. Counter was made to specifically solve this exact problem

– CoryKramer
Nov 20 '18 at 17:55

@TeraBaapBC Dictionaries in Python have no inherent order, so there is no concept of one key being "ahead" of another

– CoryKramer
Nov 20 '18 at 17:56

Isn't using d.get(item, 0) and adding it with Counter() of that item in list more efficient than doing Counter() twice?

– Austin
Nov 20 '18 at 17:58

add a comment |

This can easily be done with simple list looping and dict.get method, though other efficient method exist.

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

dict = {'orange': 4, 'apple':2, 'pears': 1}



for st in lst:

     dict[st] = dict.get(st,0)+1



dict

{'orange': 5, 'apple': 3, 'pears': 3, 'banana': 1}

edited Nov 20 '18 at 17:45

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

add a comment |

Before extending, there is a typo in line 2, which should be for string in list.

This is my suggested solution. As you iterate through the list, check each entry to see if it is a key in the dictionary (as you have already done). If it is, the dict[string] will be the number value paired with that key and you can add one to it. If not, then you can add the string as a new key with a value of 1.

# original data

lst = ['apple', 'orange', 'pears', 'pears', 'banana']  

dict = {'orange': 4, 'apple':2, 'pears': 1}



# iterate through lst and add 1 to each corresponding key value

for string in lst:

    if string in dict.keys():

        #  increment count for a found key 

        #  which can be accessed in dict[string] - no need for num

        count = int(dict[string])

        dict[string] = count + 1

    else:

        # add new key and a count of 1 to dict

        dict[string] = 1

answered Nov 20 '18 at 18:21

karsco

113

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53398545%2fcount-the-number-of-words-in-dictionary-from-a-list%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

Your answer was almost correct:

for string in lst:

    if string in dict.keys():

        dict[string] += 1

    else:

        dict[string] = 1

This is assuming that a string you haven't seen yet starts with a value of 1, which seems to be the case given your output.

You could also remove the .keys(), as python will automatically check in the keys for the values you are looping on, hence:

for string in lst:

    if string in dict:

        dict[string] += 1

    else:

        dict[string] = 1

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

add a comment |

Your answer was almost correct:

for string in lst:

    if string in dict.keys():

        dict[string] += 1

    else:

        dict[string] = 1

This is assuming that a string you haven't seen yet starts with a value of 1, which seems to be the case given your output.

You could also remove the .keys(), as python will automatically check in the keys for the values you are looping on, hence:

for string in lst:

    if string in dict:

        dict[string] += 1

    else:

        dict[string] = 1

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

add a comment |

Your answer was almost correct:

for string in lst:

    if string in dict.keys():

        dict[string] += 1

    else:

        dict[string] = 1

This is assuming that a string you haven't seen yet starts with a value of 1, which seems to be the case given your output.

You could also remove the .keys(), as python will automatically check in the keys for the values you are looping on, hence:

for string in lst:

    if string in dict:

        dict[string] += 1

    else:

        dict[string] = 1

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

Your answer was almost correct:

for string in lst:

    if string in dict.keys():

        dict[string] += 1

    else:

        dict[string] = 1

This is assuming that a string you haven't seen yet starts with a value of 1, which seems to be the case given your output.

You could also remove the .keys(), as python will automatically check in the keys for the values you are looping on, hence:

for string in lst:

    if string in dict:

        dict[string] += 1

    else:

        dict[string] = 1

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

answered Nov 20 '18 at 17:44

Luca Giorgi

179417

add a comment |

You can use collections.Counter

from collections import Counter



>>> lst = ['apple', 'orange', 'pears', 'pears', 'banana']

>>> d = {'orange': 4, 'apple':2, 'pears': 1}



>>> count = Counter(d)

>>> count

Counter({'orange': 4, 'apple': 2, 'pears': 1})

>>> count += Counter(lst)

>>> count

Counter({'orange': 5, 'pears': 3, 'apple': 3, 'banana': 1})

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

I know this is off topic, but how did it place pears ahead of apple?

– Sandesh34
Nov 20 '18 at 17:44

I don't think there is a need of Counter(d), because d itself is a dictionary of counts and you can access with d.get().

– Austin
Nov 20 '18 at 17:47

@Austin Yes but then you still have to write the update logic yourself. Counter was made to specifically solve this exact problem

– CoryKramer
Nov 20 '18 at 17:55

@TeraBaapBC Dictionaries in Python have no inherent order, so there is no concept of one key being "ahead" of another

– CoryKramer
Nov 20 '18 at 17:56

Isn't using d.get(item, 0) and adding it with Counter() of that item in list more efficient than doing Counter() twice?

– Austin
Nov 20 '18 at 17:58

add a comment |

You can use collections.Counter

from collections import Counter



>>> lst = ['apple', 'orange', 'pears', 'pears', 'banana']

>>> d = {'orange': 4, 'apple':2, 'pears': 1}



>>> count = Counter(d)

>>> count

Counter({'orange': 4, 'apple': 2, 'pears': 1})

>>> count += Counter(lst)

>>> count

Counter({'orange': 5, 'pears': 3, 'apple': 3, 'banana': 1})

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

I know this is off topic, but how did it place pears ahead of apple?

– Sandesh34
Nov 20 '18 at 17:44

I don't think there is a need of Counter(d), because d itself is a dictionary of counts and you can access with d.get().

– Austin
Nov 20 '18 at 17:47

@Austin Yes but then you still have to write the update logic yourself. Counter was made to specifically solve this exact problem

– CoryKramer
Nov 20 '18 at 17:55

@TeraBaapBC Dictionaries in Python have no inherent order, so there is no concept of one key being "ahead" of another

– CoryKramer
Nov 20 '18 at 17:56

Isn't using d.get(item, 0) and adding it with Counter() of that item in list more efficient than doing Counter() twice?

– Austin
Nov 20 '18 at 17:58

add a comment |

You can use collections.Counter

from collections import Counter



>>> lst = ['apple', 'orange', 'pears', 'pears', 'banana']

>>> d = {'orange': 4, 'apple':2, 'pears': 1}



>>> count = Counter(d)

>>> count

Counter({'orange': 4, 'apple': 2, 'pears': 1})

>>> count += Counter(lst)

>>> count

Counter({'orange': 5, 'pears': 3, 'apple': 3, 'banana': 1})

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

You can use collections.Counter

from collections import Counter



>>> lst = ['apple', 'orange', 'pears', 'pears', 'banana']

>>> d = {'orange': 4, 'apple':2, 'pears': 1}



>>> count = Counter(d)

>>> count

Counter({'orange': 4, 'apple': 2, 'pears': 1})

>>> count += Counter(lst)

>>> count

Counter({'orange': 5, 'pears': 3, 'apple': 3, 'banana': 1})

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

answered Nov 20 '18 at 17:39

CoryKramer

73.9k1188141

I know this is off topic, but how did it place pears ahead of apple?

– Sandesh34
Nov 20 '18 at 17:44

I don't think there is a need of Counter(d), because d itself is a dictionary of counts and you can access with d.get().

– Austin
Nov 20 '18 at 17:47

@Austin Yes but then you still have to write the update logic yourself. Counter was made to specifically solve this exact problem

– CoryKramer
Nov 20 '18 at 17:55

@TeraBaapBC Dictionaries in Python have no inherent order, so there is no concept of one key being "ahead" of another

– CoryKramer
Nov 20 '18 at 17:56

Isn't using d.get(item, 0) and adding it with Counter() of that item in list more efficient than doing Counter() twice?

– Austin
Nov 20 '18 at 17:58

add a comment |

I know this is off topic, but how did it place pears ahead of apple?

– Sandesh34
Nov 20 '18 at 17:44

I don't think there is a need of Counter(d), because d itself is a dictionary of counts and you can access with d.get().

– Austin
Nov 20 '18 at 17:47

@Austin Yes but then you still have to write the update logic yourself. Counter was made to specifically solve this exact problem

– CoryKramer
Nov 20 '18 at 17:55

@TeraBaapBC Dictionaries in Python have no inherent order, so there is no concept of one key being "ahead" of another

– CoryKramer
Nov 20 '18 at 17:56

Isn't using d.get(item, 0) and adding it with Counter() of that item in list more efficient than doing Counter() twice?

– Austin
Nov 20 '18 at 17:58

I know this is off topic, but how did it place pears ahead of apple?

– Sandesh34
Nov 20 '18 at 17:44

I don't think there is a need of Counter(d), because d itself is a dictionary of counts and you can access with d.get().

– Austin
Nov 20 '18 at 17:47

@Austin Yes but then you still have to write the update logic yourself. Counter was made to specifically solve this exact problem

– CoryKramer
Nov 20 '18 at 17:55

@TeraBaapBC Dictionaries in Python have no inherent order, so there is no concept of one key being "ahead" of another

– CoryKramer
Nov 20 '18 at 17:56

Isn't using d.get(item, 0) and adding it with Counter() of that item in list more efficient than doing Counter() twice?

– Austin
Nov 20 '18 at 17:58

add a comment |

This can easily be done with simple list looping and dict.get method, though other efficient method exist.

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

dict = {'orange': 4, 'apple':2, 'pears': 1}



for st in lst:

     dict[st] = dict.get(st,0)+1



dict

{'orange': 5, 'apple': 3, 'pears': 3, 'banana': 1}

edited Nov 20 '18 at 17:45

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

add a comment |

This can easily be done with simple list looping and dict.get method, though other efficient method exist.

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

dict = {'orange': 4, 'apple':2, 'pears': 1}



for st in lst:

     dict[st] = dict.get(st,0)+1



dict

{'orange': 5, 'apple': 3, 'pears': 3, 'banana': 1}

edited Nov 20 '18 at 17:45

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

add a comment |

This can easily be done with simple list looping and dict.get method, though other efficient method exist.

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

dict = {'orange': 4, 'apple':2, 'pears': 1}



for st in lst:

     dict[st] = dict.get(st,0)+1



dict

{'orange': 5, 'apple': 3, 'pears': 3, 'banana': 1}

edited Nov 20 '18 at 17:45

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

This can easily be done with simple list looping and dict.get method, though other efficient method exist.

lst = ['apple', 'orange', 'pears', 'pears', 'banana']

dict = {'orange': 4, 'apple':2, 'pears': 1}



for st in lst:

     dict[st] = dict.get(st,0)+1



dict

{'orange': 5, 'apple': 3, 'pears': 3, 'banana': 1}

edited Nov 20 '18 at 17:45

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

edited Nov 20 '18 at 17:45

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

answered Nov 20 '18 at 17:39

Rahul Chawla

551512

add a comment |

Before extending, there is a typo in line 2, which should be for string in list.

# original data

lst = ['apple', 'orange', 'pears', 'pears', 'banana']  

dict = {'orange': 4, 'apple':2, 'pears': 1}



# iterate through lst and add 1 to each corresponding key value

for string in lst:

    if string in dict.keys():

        #  increment count for a found key 

        #  which can be accessed in dict[string] - no need for num

        count = int(dict[string])

        dict[string] = count + 1

    else:

        # add new key and a count of 1 to dict

        dict[string] = 1

answered Nov 20 '18 at 18:21

karsco

113

add a comment |

Before extending, there is a typo in line 2, which should be for string in list.

# original data

lst = ['apple', 'orange', 'pears', 'pears', 'banana']  

dict = {'orange': 4, 'apple':2, 'pears': 1}



# iterate through lst and add 1 to each corresponding key value

for string in lst:

    if string in dict.keys():

        #  increment count for a found key 

        #  which can be accessed in dict[string] - no need for num

        count = int(dict[string])

        dict[string] = count + 1

    else:

        # add new key and a count of 1 to dict

        dict[string] = 1

answered Nov 20 '18 at 18:21

karsco

113

add a comment |

Before extending, there is a typo in line 2, which should be for string in list.

# original data

lst = ['apple', 'orange', 'pears', 'pears', 'banana']  

dict = {'orange': 4, 'apple':2, 'pears': 1}



# iterate through lst and add 1 to each corresponding key value

for string in lst:

    if string in dict.keys():

        #  increment count for a found key 

        #  which can be accessed in dict[string] - no need for num

        count = int(dict[string])

        dict[string] = count + 1

    else:

        # add new key and a count of 1 to dict

        dict[string] = 1

answered Nov 20 '18 at 18:21

karsco

113

Before extending, there is a typo in line 2, which should be for string in list.

# original data

lst = ['apple', 'orange', 'pears', 'pears', 'banana']  

dict = {'orange': 4, 'apple':2, 'pears': 1}



# iterate through lst and add 1 to each corresponding key value

for string in lst:

    if string in dict.keys():

        #  increment count for a found key 

        #  which can be accessed in dict[string] - no need for num

        count = int(dict[string])

        dict[string] = count + 1

    else:

        # add new key and a count of 1 to dict

        dict[string] = 1

answered Nov 20 '18 at 18:21

karsco

113

answered Nov 20 '18 at 18:21

karsco

113

answered Nov 20 '18 at 18:21

karsco

113

answered Nov 20 '18 at 18:21

karsco

113

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

Search This Blog

Ufyukyu