Filter and Generate Sub-DataFrame From a List





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}







0















I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:



sales=[{'name':'Samsung Korea','model':'S1'},

   {'name':'Samsung Vienam','model':'J1'},

   {'name':'LG America','model':'L1'}

  ]

df=pd.DataFrame(sales)


I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are



dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}] 



dataLG = [{'name': 'LG', 'model': 'L1'}]


I tried:



 customer={}

 for i in companyList:

       customer[i] = df[df.name.str.contains('i')]


but this gives me a wrong answer. Could you help me to fix this case?






python pandas dataframe







share|improve this question

























  • Thats a set not a list!

    – U9-Forward
    Nov 25 '18 at 0:22











  • This is a typo. Use i not 'i', the latter is just a string, the former references an element in companyList.

    – jpp
    Nov 25 '18 at 1:37













  • it works after fixing, thanks @jpp

    – HTB
    Nov 26 '18 at 18:26


















0















I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:



sales=[{'name':'Samsung Korea','model':'S1'},

   {'name':'Samsung Vienam','model':'J1'},

   {'name':'LG America','model':'L1'}

  ]

df=pd.DataFrame(sales)


I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are



dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}] 



dataLG = [{'name': 'LG', 'model': 'L1'}]


I tried:



 customer={}

 for i in companyList:

       customer[i] = df[df.name.str.contains('i')]


but this gives me a wrong answer. Could you help me to fix this case?






python pandas dataframe







share|improve this question

























  • Thats a set not a list!

    – U9-Forward
    Nov 25 '18 at 0:22











  • This is a typo. Use i not 'i', the latter is just a string, the former references an element in companyList.

    – jpp
    Nov 25 '18 at 1:37













  • it works after fixing, thanks @jpp

    – HTB
    Nov 26 '18 at 18:26














0












0








0








I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:



sales=[{'name':'Samsung Korea','model':'S1'},

   {'name':'Samsung Vienam','model':'J1'},

   {'name':'LG America','model':'L1'}

  ]

df=pd.DataFrame(sales)


I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are



dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}] 



dataLG = [{'name': 'LG', 'model': 'L1'}]


I tried:



 customer={}

 for i in companyList:

       customer[i] = df[df.name.str.contains('i')]


but this gives me a wrong answer. Could you help me to fix this case?






python pandas dataframe







share|improve this question
















I have a set:
CompanyList={'Apple','LG','Samsung'}
and a pandas DataFrame:



sales=[{'name':'Samsung Korea','model':'S1'},

   {'name':'Samsung Vienam','model':'J1'},

   {'name':'LG America','model':'L1'}

  ]

df=pd.DataFrame(sales)


I'd like to go through the CompanyList, then generate new Sub-DataFrame from 'sales' DataFrame. The expected results are



dataSamsung = [{'name': 'Samsung', 'model': 'S1'},{'name': 'Samsung', 'model': 'J1'}] 



dataLG = [{'name': 'LG', 'model': 'L1'}]


I tried:



 customer={}

 for i in companyList:

       customer[i] = df[df.name.str.contains('i')]


but this gives me a wrong answer. Could you help me to fix this case?






python pandas dataframe




python pandas dataframe






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 25 '18 at 1:32









U9-Forward

18.6k51744




18.6k51744










asked Nov 25 '18 at 0:22









HTBHTB

41




41













  • Thats a set not a list!

    – U9-Forward
    Nov 25 '18 at 0:22











  • This is a typo. Use i not 'i', the latter is just a string, the former references an element in companyList.

    – jpp
    Nov 25 '18 at 1:37













  • it works after fixing, thanks @jpp

    – HTB
    Nov 26 '18 at 18:26



















  • Thats a set not a list!

    – U9-Forward
    Nov 25 '18 at 0:22











  • This is a typo. Use i not 'i', the latter is just a string, the former references an element in companyList.

    – jpp
    Nov 25 '18 at 1:37













  • it works after fixing, thanks @jpp

    – HTB
    Nov 26 '18 at 18:26

















Thats a set not a list!

– U9-Forward
Nov 25 '18 at 0:22





Thats a set not a list!

– U9-Forward
Nov 25 '18 at 0:22













This is a typo. Use i not 'i', the latter is just a string, the former references an element in companyList.

– jpp
Nov 25 '18 at 1:37







This is a typo. Use i not 'i', the latter is just a string, the former references an element in companyList.

– jpp
Nov 25 '18 at 1:37















it works after fixing, thanks @jpp

– HTB
Nov 26 '18 at 18:26





it works after fixing, thanks @jpp

– HTB
Nov 26 '18 at 18:26












1 Answer
1






active

oldest

votes


















0














Try apply:



df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])


apply with list comprehension.






share|improve this answer
























  • Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

    – HTB
    Nov 26 '18 at 18:03












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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53463609%2ffilter-and-generate-sub-dataframe-from-a-list%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0














Try apply:



df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])


apply with list comprehension.






share|improve this answer
























  • Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

    – HTB
    Nov 26 '18 at 18:03
















0














Try apply:



df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])


apply with list comprehension.






share|improve this answer
























  • Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

    – HTB
    Nov 26 '18 at 18:03














0












0








0







Try apply:



df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])


apply with list comprehension.






share|improve this answer













Try apply:



df['name']=df['name'].apply(lambda x: [i for i in CompanyList if i in x][0])


apply with list comprehension.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 25 '18 at 0:26









U9-ForwardU9-Forward

18.6k51744




18.6k51744













  • Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

    – HTB
    Nov 26 '18 at 18:03



















  • Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

    – HTB
    Nov 26 '18 at 18:03

















Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

– HTB
Nov 26 '18 at 18:03





Thanks for your help, but can we separate the original DataFrame into smaller ones based on the name?

– HTB
Nov 26 '18 at 18:03




















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














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53463609%2ffilter-and-generate-sub-dataframe-from-a-list%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

這個網誌中的熱門文章

Tangent Lines Diagram Along Smooth Curve

圖片
7 1 I am trying to replicate a diagram and have a minimal example. I don't know how to add additional tangents between the specified points and preserve smoothness. documentclass{article} usepackage{tikz} usetikzlibrary{calc,intersections} begin{document} begin{tikzpicture} % Axes draw [->, name path=x] (-1,0) -- (11,0) node [right] {$x$}; draw [->] (0,-1) -- (0,6) node [above] {$y$}; % Origin node at (0,0) [below left] {$0$}; % Points coordinate (start) at (1,-0.8); coordinate (c1) at (3,3); coordinate (c2) at (5.5,1.5); coordinate (c3) at (8,4); coordinate (end) at (10.5,-0.8); % show the points foreach n in {start,c1,c2,c3,end} fill [black] (n) circle (2pt) node [below] {}; % join the coordinates draw [thick,name path=curve] (start) to[out=70,in=180] (c1) to[out=0,in=180]
閱讀完整內容

Yusuf al-Mu'taman ibn Hud

圖片
Main article: Taifa of Zaragoza Al-Mu'taman Ruler of Zaragoza Reign 1081–1085 Predecessor Ahmad al-Muqtadir Successor Al-Mustain II Born Zaragoza Died 1085 Full name Yusuf ibn Ahmad al-Mu'taman ibn Hūd House Banu Hud Yusuf ibn Ahmad al-Mu'taman ibn Hūd (Arabic: المؤتمن بالله يوسف إبن أحمد إبن هود ‎, al-Mutaman bi l-Lah , died c. 1085) was the third king of the Banu Hud dynasty who ruled over the Taifa of Zaragoza from 1081 to 1085. Yusuf al-Mutaman reigned during the height of power of Muslim Zaragoza , following the thriving period of his father Ahmad al-Muqtadir. He continued his father's efforts and created around him a court of intellectuals, living in the beautiful palace of Aljafería, nicknamed as "the palace of joy". Al-Mu'taman was also a scholarly king and a patron of science, philosophy and arts. As an example of a wise king, he knew astrology, philosophy, and especially mathematics, a discipline in which
閱讀完整內容

Zucchini

圖片
This article is about the vegetable. For other uses, see Zucchini (disambiguation). Zucchini (courgette) A striped and a uniform color zucchini Genus Cucurbita Species Cucurbita pepo Origin 19th century northern Italy The zucchini ( / z uː ˈ k iː n i / , American English) or courgette ( / k ʊər ˈ ʒ ɛ t / , British English) is a summer squash which can reach nearly 1 metre (100 cm; 39 in) in length, but is usually harvested when still immature at about 15 to 25 cm (6 to 10 in). [1] A zucchini is a thin-skinned cultivar of what in Britain and Ireland is referred to as a marrow. [2] [3] In South Africa, a zucchini is known as a baby marrow. Along with certain other squashes and pumpkins, the zucchini belongs to the species Cucurbita pepo . It can be dark or light green. A related hybrid, the golden zucchini, is a deep yellow or orange color. [4] In a culinary context, the zucchini is treated as a vegetable; it is usually cooked and presented as a savory dish o
閱讀完整內容