change pandas df to python dict
0
I would like to change the following pandas DataFrame object to a python dictionary, and then I want to get the value with energy_emob_BB and energy_emob_BE.
emob_energy = pd.DataFrame(session.query(abbb_emob.id,
abbb_emob.scenario,
abbb_emob.region,
abbb_emob.energy).filter(
abbb_emob.scenario == 'ES2030').all())
energy_emob_BB = float(emob_energy.query('region=="BB"')['energy'])
energy_emob_BE = float(emob_energy.query('region=="BE"')['energy'])
Here is some output of the DataFrame and 2 objects which I have to get from the DataFrame
(Pdb) emob_energy
id scenario region energy
0 1 ES2030 BB 2183000.0
1 2 ES2030 BE 1298333.0
(Pdb) energy_emob_BB
2183000.0
(Pdb) energy_emob_BE
1298333.0
How would I make .query to work with python dict?
python pandas dictionary
edited Nov 20 '18 at 16:41
user3471881
1,1922619
asked Nov 20 '18 at 13:22
oakcaoakca
334112
add a comment |
0
I would like to change the following pandas DataFrame object to a python dictionary, and then I want to get the value with energy_emob_BB and energy_emob_BE.
emob_energy = pd.DataFrame(session.query(abbb_emob.id,
abbb_emob.scenario,
abbb_emob.region,
abbb_emob.energy).filter(
abbb_emob.scenario == 'ES2030').all())
energy_emob_BB = float(emob_energy.query('region=="BB"')['energy'])
energy_emob_BE = float(emob_energy.query('region=="BE"')['energy'])
Here is some output of the DataFrame and 2 objects which I have to get from the DataFrame
(Pdb) emob_energy
id scenario region energy
0 1 ES2030 BB 2183000.0
1 2 ES2030 BE 1298333.0
(Pdb) energy_emob_BB
2183000.0
(Pdb) energy_emob_BE
1298333.0
How would I make .query to work with python dict?
python pandas dictionary
edited Nov 20 '18 at 16:41
user3471881
1,1922619
asked Nov 20 '18 at 13:22
oakcaoakca
334112
add a comment |
0
0
0
I would like to change the following pandas DataFrame object to a python dictionary, and then I want to get the value with energy_emob_BB and energy_emob_BE.
emob_energy = pd.DataFrame(session.query(abbb_emob.id,
abbb_emob.scenario,
abbb_emob.region,
abbb_emob.energy).filter(
abbb_emob.scenario == 'ES2030').all())
energy_emob_BB = float(emob_energy.query('region=="BB"')['energy'])
energy_emob_BE = float(emob_energy.query('region=="BE"')['energy'])
Here is some output of the DataFrame and 2 objects which I have to get from the DataFrame
(Pdb) emob_energy
id scenario region energy
0 1 ES2030 BB 2183000.0
1 2 ES2030 BE 1298333.0
(Pdb) energy_emob_BB
2183000.0
(Pdb) energy_emob_BE
1298333.0
How would I make .query to work with python dict?
python pandas dictionary
edited Nov 20 '18 at 16:41
user3471881
1,1922619
asked Nov 20 '18 at 13:22
oakcaoakca
334112
I would like to change the following pandas DataFrame object to a python dictionary, and then I want to get the value with energy_emob_BB and energy_emob_BE.
emob_energy = pd.DataFrame(session.query(abbb_emob.id,
abbb_emob.scenario,
abbb_emob.region,
abbb_emob.energy).filter(
abbb_emob.scenario == 'ES2030').all())
energy_emob_BB = float(emob_energy.query('region=="BB"')['energy'])
energy_emob_BE = float(emob_energy.query('region=="BE"')['energy'])
Here is some output of the DataFrame and 2 objects which I have to get from the DataFrame
(Pdb) emob_energy
id scenario region energy
0 1 ES2030 BB 2183000.0
1 2 ES2030 BE 1298333.0
(Pdb) energy_emob_BB
2183000.0
(Pdb) energy_emob_BE
1298333.0
How would I make .query to work with python dict?
python pandas dictionary
python pandas dictionary
edited Nov 20 '18 at 16:41
user3471881
1,1922619
asked Nov 20 '18 at 13:22
oakcaoakca
334112
edited Nov 20 '18 at 16:41
user3471881
1,1922619
asked Nov 20 '18 at 13:22
oakcaoakca
334112
edited Nov 20 '18 at 16:41
user3471881
1,1922619
edited Nov 20 '18 at 16:41
user3471881
1,1922619
edited Nov 20 '18 at 16:41
user3471881
1,1922619
1,1922619
asked Nov 20 '18 at 13:22
oakcaoakca
334112
asked Nov 20 '18 at 13:22
oakcaoakca
334112
asked Nov 20 '18 at 13:22
oakcaoakca
334112
334112
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
pandas.DataFrame.to_dict renders the data a dicts in dicts by default.
df_dict = emob_energy.to_dict()
print(df_dict['energy_emob_bb']) # {0: 2183000.0, 1: 1298333.0, ...}
If you want them as dict of lists:
df_dict = emob_energy.to_dict(orient='list')
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
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');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53393955%2fchange-pandas-df-to-python-dict%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
2
pandas.DataFrame.to_dict renders the data a dicts in dicts by default.
df_dict = emob_energy.to_dict()
print(df_dict['energy_emob_bb']) # {0: 2183000.0, 1: 1298333.0, ...}
If you want them as dict of lists:
df_dict = emob_energy.to_dict(orient='list')
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
add a comment |
2
pandas.DataFrame.to_dict renders the data a dicts in dicts by default.
df_dict = emob_energy.to_dict()
print(df_dict['energy_emob_bb']) # {0: 2183000.0, 1: 1298333.0, ...}
If you want them as dict of lists:
df_dict = emob_energy.to_dict(orient='list')
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
add a comment |
2
2
2
pandas.DataFrame.to_dict renders the data a dicts in dicts by default.
df_dict = emob_energy.to_dict()
print(df_dict['energy_emob_bb']) # {0: 2183000.0, 1: 1298333.0, ...}
If you want them as dict of lists:
df_dict = emob_energy.to_dict(orient='list')
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
pandas.DataFrame.to_dict renders the data a dicts in dicts by default.
df_dict = emob_energy.to_dict()
print(df_dict['energy_emob_bb']) # {0: 2183000.0, 1: 1298333.0, ...}
If you want them as dict of lists:
df_dict = emob_energy.to_dict(orient='list')
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
answered Nov 20 '18 at 13:28
Charles LandauCharles Landau
2,6381216
2,6381216
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
add a comment |
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
can you actually create the dict without creating the dataframe?
– oakca
Nov 20 '18 at 13:28
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
Of course! But I don't know how without seeing your data. If I responded to your original question then mark it as answered please.
– Charles Landau
Nov 20 '18 at 13:55
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');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53393955%2fchange-pandas-df-to-python-dict%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
