Mixing
How to count a value with JSON based on another value in the JSON file
I have a JSON file made up of an array which contains dictionaries, each dictionary is an opinion of a buyer on a specific garage.
I want to find out how many occurrence of each car's type I have in each garage, it looks like this:
[
{"garage": "mike_gar", "reliability": 6, "car_type": "ford", "time": "16:10:36"},
{"garage": "bill_gar", "reliability": 5,"car_type": "kia", "time": "4:37:22"},
{"garage": "alison_gar", "reliability": 1, "car_type": "kia", "time": "11:25:40"},
{"garage": "alison_gar", "reliability": 10, "car_type": "mazda", "time": "2:18:42"},
{"garage": "mike_gar", "reliability": 3, "car_type": "mazda", "time": "12:14:20"},
{"garage": "mike_gar", "reliability": 2, "car_type": "ford", "time": "2:08:27"}
]
Assuming we already read from the JSON file to a variable g_arr.
I'v tried to use reduce() to count the occurrence , but failed to succeed.
output example: {"garage" : "mike_gar", "types":{"ford" : 2, "mazda": 1}}
python json dictionary
edited Dec 31 '18 at 21:37
Jack Moody
7481723
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
add a comment |
I have a JSON file made up of an array which contains dictionaries, each dictionary is an opinion of a buyer on a specific garage.
I want to find out how many occurrence of each car's type I have in each garage, it looks like this:
[
{"garage": "mike_gar", "reliability": 6, "car_type": "ford", "time": "16:10:36"},
{"garage": "bill_gar", "reliability": 5,"car_type": "kia", "time": "4:37:22"},
{"garage": "alison_gar", "reliability": 1, "car_type": "kia", "time": "11:25:40"},
{"garage": "alison_gar", "reliability": 10, "car_type": "mazda", "time": "2:18:42"},
{"garage": "mike_gar", "reliability": 3, "car_type": "mazda", "time": "12:14:20"},
{"garage": "mike_gar", "reliability": 2, "car_type": "ford", "time": "2:08:27"}
]
Assuming we already read from the JSON file to a variable g_arr.
I'v tried to use reduce() to count the occurrence , but failed to succeed.
output example: {"garage" : "mike_gar", "types":{"ford" : 2, "mazda": 1}}
python json dictionary
edited Dec 31 '18 at 21:37
Jack Moody
7481723
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
1
Is Jason the guy who owns the garage?
– wim
Dec 31 '18 at 21:32
@wim I think it's supposed to be JSON. Fixed that.
– Jack Moody
Dec 31 '18 at 21:37
@wim opppsss...
– Shelly875
Dec 31 '18 at 21:38
@Shelly875 I think you will find this answer helpful. Let us know if that solves your problem. It isn't an exact duplicate, but it is pretty closely related.
– Jack Moody
Dec 31 '18 at 21:39
add a comment |
I have a JSON file made up of an array which contains dictionaries, each dictionary is an opinion of a buyer on a specific garage.
I want to find out how many occurrence of each car's type I have in each garage, it looks like this:
[
{"garage": "mike_gar", "reliability": 6, "car_type": "ford", "time": "16:10:36"},
{"garage": "bill_gar", "reliability": 5,"car_type": "kia", "time": "4:37:22"},
{"garage": "alison_gar", "reliability": 1, "car_type": "kia", "time": "11:25:40"},
{"garage": "alison_gar", "reliability": 10, "car_type": "mazda", "time": "2:18:42"},
{"garage": "mike_gar", "reliability": 3, "car_type": "mazda", "time": "12:14:20"},
{"garage": "mike_gar", "reliability": 2, "car_type": "ford", "time": "2:08:27"}
]
Assuming we already read from the JSON file to a variable g_arr.
I'v tried to use reduce() to count the occurrence , but failed to succeed.
output example: {"garage" : "mike_gar", "types":{"ford" : 2, "mazda": 1}}
python json dictionary
edited Dec 31 '18 at 21:37
Jack Moody
7481723
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
I have a JSON file made up of an array which contains dictionaries, each dictionary is an opinion of a buyer on a specific garage.
I want to find out how many occurrence of each car's type I have in each garage, it looks like this:
[
{"garage": "mike_gar", "reliability": 6, "car_type": "ford", "time": "16:10:36"},
{"garage": "bill_gar", "reliability": 5,"car_type": "kia", "time": "4:37:22"},
{"garage": "alison_gar", "reliability": 1, "car_type": "kia", "time": "11:25:40"},
{"garage": "alison_gar", "reliability": 10, "car_type": "mazda", "time": "2:18:42"},
{"garage": "mike_gar", "reliability": 3, "car_type": "mazda", "time": "12:14:20"},
{"garage": "mike_gar", "reliability": 2, "car_type": "ford", "time": "2:08:27"}
]
Assuming we already read from the JSON file to a variable g_arr.
I'v tried to use reduce() to count the occurrence , but failed to succeed.
output example: {"garage" : "mike_gar", "types":{"ford" : 2, "mazda": 1}}
python json dictionary
python json dictionary
edited Dec 31 '18 at 21:37
Jack Moody
7481723
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
edited Dec 31 '18 at 21:37
Jack Moody
7481723
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
edited Dec 31 '18 at 21:37
Jack Moody
7481723
edited Dec 31 '18 at 21:37
Jack Moody
7481723
edited Dec 31 '18 at 21:37
Jack Moody
7481723
7481723
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
asked Dec 31 '18 at 21:26
Shelly875Shelly875
376
376
1
Is Jason the guy who owns the garage?
– wim
Dec 31 '18 at 21:32
@wim I think it's supposed to be JSON. Fixed that.
– Jack Moody
Dec 31 '18 at 21:37
@wim opppsss...
– Shelly875
Dec 31 '18 at 21:38
@Shelly875 I think you will find this answer helpful. Let us know if that solves your problem. It isn't an exact duplicate, but it is pretty closely related.
– Jack Moody
Dec 31 '18 at 21:39
add a comment |
1
Is Jason the guy who owns the garage?
– wim
Dec 31 '18 at 21:32
@wim I think it's supposed to be JSON. Fixed that.
– Jack Moody
Dec 31 '18 at 21:37
@wim opppsss...
– Shelly875
Dec 31 '18 at 21:38
@Shelly875 I think you will find this answer helpful. Let us know if that solves your problem. It isn't an exact duplicate, but it is pretty closely related.
– Jack Moody
Dec 31 '18 at 21:39
1
1
Is Jason the guy who owns the garage?
– wim
Dec 31 '18 at 21:32
Is Jason the guy who owns the garage?
– wim
Dec 31 '18 at 21:32
@wim I think it's supposed to be JSON. Fixed that.
– Jack Moody
Dec 31 '18 at 21:37
@wim I think it's supposed to be JSON. Fixed that.
– Jack Moody
Dec 31 '18 at 21:37
@wim opppsss...
– Shelly875
Dec 31 '18 at 21:38
@wim opppsss...
– Shelly875
Dec 31 '18 at 21:38
@Shelly875 I think you will find this answer helpful. Let us know if that solves your problem. It isn't an exact duplicate, but it is pretty closely related.
– Jack Moody
Dec 31 '18 at 21:39
@Shelly875 I think you will find this answer helpful. Let us know if that solves your problem. It isn't an exact duplicate, but it is pretty closely related.
– Jack Moody
Dec 31 '18 at 21:39
add a comment |
3 Answers
3
active
oldest
votes
Here's a solution based on reduction. First, I test whether the garage exists in the accumulation dictionary, and if not, create it. Then, I check whether the car type exists in the garage dictionary, and if not, I create it. Finally, I increment the car type.
res = {}
for d in garages:
if d["garage"] not in res:
res[d["garage"]] = {"garage": d["garage"], "types": {}}
if d["car_type"] not in res[d["garage"]]["types"]:
res[d["garage"]]["types"][d["car_type"]] = 0
res[d["garage"]]["types"][d["car_type"]] += 1
Output:
{
'mike_gar': {'garage': 'mike_gar', 'types': {'ford': 2, 'mazda': 1}},
'bill_gar': {'garage': 'bill_gar', 'types': {'kia': 1}},
'alison_gar': {'garage': 'alison_gar', 'types': {'kia': 1, 'mazda': 1}}
}
Try it!
If you'd like your result in an array, use res.values().
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
1
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
add a comment |
You can simply parse your data and do the count as the following:
garages =
cars =
output =
for element in data:
if element['garage'] not in garages: garages.append(element['garage'])
if element['car_type'] not in cars: cars.append(element['car_type'])
for type in garages:
current = {}
current['types'] = {}
current['garage'] = type
for element in data:
if element['car_type'] not in current['types']:
current['types'][element['car_type']]=0
if current['garage'] == element['garage']:
for car_type in cars:
if element['car_type'] == car_type:
current['types'][element['car_type']]+=1
output.append(current)
print output
the output of executing the above is:
[{'garage': 'mike_gar', 'types': {'mazda': 1, 'kia': 0, 'ford': 2}}, {'garage': 'bill_gar', 'types': {'mazda': 0, 'kia': 1, 'ford': 0}}, {'garage': 'alison_gar', 'types': {'mazda': 1, 'kia': 1, 'ford': 0}}]
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
add a comment |
Pandas package is great for working with a such data. You can easily convert your list into a Pandas dataframe.
import pandas as pd
df = pd.DataFrame(g_arr)
print(df)
Prints:
car_type garage reliability time
0 ford mike_gar 6 16:10:36
1 kia bill_gar 5 4:37:22
2 kia alison_gar 1 11:25:40
3 mazda alison_gar 10 2:18:42
4 mazda mike_gar 3 12:14:20
5 ford mike_gar 2 2:08:27
Than you can use the .groupby() method to group your data and the .size() method to get row counts per group.
print(df.groupby(['garage', 'car_type']).size())
Prints:
garage car_type
alison_gar kia 1
mazda 1
bill_gar kia 1
mike_gar ford 2
mazda 1
dtype: int64
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
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
});
}
});
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%2f53991537%2fhow-to-count-a-value-with-json-based-on-another-value-in-the-json-file%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's a solution based on reduction. First, I test whether the garage exists in the accumulation dictionary, and if not, create it. Then, I check whether the car type exists in the garage dictionary, and if not, I create it. Finally, I increment the car type.
res = {}
for d in garages:
if d["garage"] not in res:
res[d["garage"]] = {"garage": d["garage"], "types": {}}
if d["car_type"] not in res[d["garage"]]["types"]:
res[d["garage"]]["types"][d["car_type"]] = 0
res[d["garage"]]["types"][d["car_type"]] += 1
Output:
{
'mike_gar': {'garage': 'mike_gar', 'types': {'ford': 2, 'mazda': 1}},
'bill_gar': {'garage': 'bill_gar', 'types': {'kia': 1}},
'alison_gar': {'garage': 'alison_gar', 'types': {'kia': 1, 'mazda': 1}}
}
Try it!
If you'd like your result in an array, use res.values().
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
1
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
add a comment |
Here's a solution based on reduction. First, I test whether the garage exists in the accumulation dictionary, and if not, create it. Then, I check whether the car type exists in the garage dictionary, and if not, I create it. Finally, I increment the car type.
res = {}
for d in garages:
if d["garage"] not in res:
res[d["garage"]] = {"garage": d["garage"], "types": {}}
if d["car_type"] not in res[d["garage"]]["types"]:
res[d["garage"]]["types"][d["car_type"]] = 0
res[d["garage"]]["types"][d["car_type"]] += 1
Output:
{
'mike_gar': {'garage': 'mike_gar', 'types': {'ford': 2, 'mazda': 1}},
'bill_gar': {'garage': 'bill_gar', 'types': {'kia': 1}},
'alison_gar': {'garage': 'alison_gar', 'types': {'kia': 1, 'mazda': 1}}
}
Try it!
If you'd like your result in an array, use res.values().
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
1
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
add a comment |
Here's a solution based on reduction. First, I test whether the garage exists in the accumulation dictionary, and if not, create it. Then, I check whether the car type exists in the garage dictionary, and if not, I create it. Finally, I increment the car type.
res = {}
for d in garages:
if d["garage"] not in res:
res[d["garage"]] = {"garage": d["garage"], "types": {}}
if d["car_type"] not in res[d["garage"]]["types"]:
res[d["garage"]]["types"][d["car_type"]] = 0
res[d["garage"]]["types"][d["car_type"]] += 1
Output:
{
'mike_gar': {'garage': 'mike_gar', 'types': {'ford': 2, 'mazda': 1}},
'bill_gar': {'garage': 'bill_gar', 'types': {'kia': 1}},
'alison_gar': {'garage': 'alison_gar', 'types': {'kia': 1, 'mazda': 1}}
}
Try it!
If you'd like your result in an array, use res.values().
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
Here's a solution based on reduction. First, I test whether the garage exists in the accumulation dictionary, and if not, create it. Then, I check whether the car type exists in the garage dictionary, and if not, I create it. Finally, I increment the car type.
res = {}
for d in garages:
if d["garage"] not in res:
res[d["garage"]] = {"garage": d["garage"], "types": {}}
if d["car_type"] not in res[d["garage"]]["types"]:
res[d["garage"]]["types"][d["car_type"]] = 0
res[d["garage"]]["types"][d["car_type"]] += 1
Output:
{
'mike_gar': {'garage': 'mike_gar', 'types': {'ford': 2, 'mazda': 1}},
'bill_gar': {'garage': 'bill_gar', 'types': {'kia': 1}},
'alison_gar': {'garage': 'alison_gar', 'types': {'kia': 1, 'mazda': 1}}
}
Try it!
If you'd like your result in an array, use res.values().
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
edited Dec 31 '18 at 22:02
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
answered Dec 31 '18 at 21:51
ggorlenggorlen
7,3883826
7,3883826
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
1
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
add a comment |
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
1
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
That is a Javascript solution, the question has a Python tag.
– amanb
Dec 31 '18 at 21:55
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
Whoops--coded before looking! Changed to Python.
– ggorlen
Dec 31 '18 at 22:03
1
1
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
I'v tried the code and that's exactly what I was looking for! thank you :)
– Shelly875
Jan 1 at 9:21
add a comment |
You can simply parse your data and do the count as the following:
garages =
cars =
output =
for element in data:
if element['garage'] not in garages: garages.append(element['garage'])
if element['car_type'] not in cars: cars.append(element['car_type'])
for type in garages:
current = {}
current['types'] = {}
current['garage'] = type
for element in data:
if element['car_type'] not in current['types']:
current['types'][element['car_type']]=0
if current['garage'] == element['garage']:
for car_type in cars:
if element['car_type'] == car_type:
current['types'][element['car_type']]+=1
output.append(current)
print output
the output of executing the above is:
[{'garage': 'mike_gar', 'types': {'mazda': 1, 'kia': 0, 'ford': 2}}, {'garage': 'bill_gar', 'types': {'mazda': 0, 'kia': 1, 'ford': 0}}, {'garage': 'alison_gar', 'types': {'mazda': 1, 'kia': 1, 'ford': 0}}]
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
add a comment |
You can simply parse your data and do the count as the following:
garages =
cars =
output =
for element in data:
if element['garage'] not in garages: garages.append(element['garage'])
if element['car_type'] not in cars: cars.append(element['car_type'])
for type in garages:
current = {}
current['types'] = {}
current['garage'] = type
for element in data:
if element['car_type'] not in current['types']:
current['types'][element['car_type']]=0
if current['garage'] == element['garage']:
for car_type in cars:
if element['car_type'] == car_type:
current['types'][element['car_type']]+=1
output.append(current)
print output
the output of executing the above is:
[{'garage': 'mike_gar', 'types': {'mazda': 1, 'kia': 0, 'ford': 2}}, {'garage': 'bill_gar', 'types': {'mazda': 0, 'kia': 1, 'ford': 0}}, {'garage': 'alison_gar', 'types': {'mazda': 1, 'kia': 1, 'ford': 0}}]
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
add a comment |
You can simply parse your data and do the count as the following:
garages =
cars =
output =
for element in data:
if element['garage'] not in garages: garages.append(element['garage'])
if element['car_type'] not in cars: cars.append(element['car_type'])
for type in garages:
current = {}
current['types'] = {}
current['garage'] = type
for element in data:
if element['car_type'] not in current['types']:
current['types'][element['car_type']]=0
if current['garage'] == element['garage']:
for car_type in cars:
if element['car_type'] == car_type:
current['types'][element['car_type']]+=1
output.append(current)
print output
the output of executing the above is:
[{'garage': 'mike_gar', 'types': {'mazda': 1, 'kia': 0, 'ford': 2}}, {'garage': 'bill_gar', 'types': {'mazda': 0, 'kia': 1, 'ford': 0}}, {'garage': 'alison_gar', 'types': {'mazda': 1, 'kia': 1, 'ford': 0}}]
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
You can simply parse your data and do the count as the following:
garages =
cars =
output =
for element in data:
if element['garage'] not in garages: garages.append(element['garage'])
if element['car_type'] not in cars: cars.append(element['car_type'])
for type in garages:
current = {}
current['types'] = {}
current['garage'] = type
for element in data:
if element['car_type'] not in current['types']:
current['types'][element['car_type']]=0
if current['garage'] == element['garage']:
for car_type in cars:
if element['car_type'] == car_type:
current['types'][element['car_type']]+=1
output.append(current)
print output
the output of executing the above is:
[{'garage': 'mike_gar', 'types': {'mazda': 1, 'kia': 0, 'ford': 2}}, {'garage': 'bill_gar', 'types': {'mazda': 0, 'kia': 1, 'ford': 0}}, {'garage': 'alison_gar', 'types': {'mazda': 1, 'kia': 1, 'ford': 0}}]
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
answered Dec 31 '18 at 21:57
Walid Da.Walid Da.
6041413
6041413
add a comment |
add a comment |
Pandas package is great for working with a such data. You can easily convert your list into a Pandas dataframe.
import pandas as pd
df = pd.DataFrame(g_arr)
print(df)
Prints:
car_type garage reliability time
0 ford mike_gar 6 16:10:36
1 kia bill_gar 5 4:37:22
2 kia alison_gar 1 11:25:40
3 mazda alison_gar 10 2:18:42
4 mazda mike_gar 3 12:14:20
5 ford mike_gar 2 2:08:27
Than you can use the .groupby() method to group your data and the .size() method to get row counts per group.
print(df.groupby(['garage', 'car_type']).size())
Prints:
garage car_type
alison_gar kia 1
mazda 1
bill_gar kia 1
mike_gar ford 2
mazda 1
dtype: int64
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
add a comment |
Pandas package is great for working with a such data. You can easily convert your list into a Pandas dataframe.
import pandas as pd
df = pd.DataFrame(g_arr)
print(df)
Prints:
car_type garage reliability time
0 ford mike_gar 6 16:10:36
1 kia bill_gar 5 4:37:22
2 kia alison_gar 1 11:25:40
3 mazda alison_gar 10 2:18:42
4 mazda mike_gar 3 12:14:20
5 ford mike_gar 2 2:08:27
Than you can use the .groupby() method to group your data and the .size() method to get row counts per group.
print(df.groupby(['garage', 'car_type']).size())
Prints:
garage car_type
alison_gar kia 1
mazda 1
bill_gar kia 1
mike_gar ford 2
mazda 1
dtype: int64
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
add a comment |
Pandas package is great for working with a such data. You can easily convert your list into a Pandas dataframe.
import pandas as pd
df = pd.DataFrame(g_arr)
print(df)
Prints:
car_type garage reliability time
0 ford mike_gar 6 16:10:36
1 kia bill_gar 5 4:37:22
2 kia alison_gar 1 11:25:40
3 mazda alison_gar 10 2:18:42
4 mazda mike_gar 3 12:14:20
5 ford mike_gar 2 2:08:27
Than you can use the .groupby() method to group your data and the .size() method to get row counts per group.
print(df.groupby(['garage', 'car_type']).size())
Prints:
garage car_type
alison_gar kia 1
mazda 1
bill_gar kia 1
mike_gar ford 2
mazda 1
dtype: int64
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
Pandas package is great for working with a such data. You can easily convert your list into a Pandas dataframe.
import pandas as pd
df = pd.DataFrame(g_arr)
print(df)
Prints:
car_type garage reliability time
0 ford mike_gar 6 16:10:36
1 kia bill_gar 5 4:37:22
2 kia alison_gar 1 11:25:40
3 mazda alison_gar 10 2:18:42
4 mazda mike_gar 3 12:14:20
5 ford mike_gar 2 2:08:27
Than you can use the .groupby() method to group your data and the .size() method to get row counts per group.
print(df.groupby(['garage', 'car_type']).size())
Prints:
garage car_type
alison_gar kia 1
mazda 1
bill_gar kia 1
mike_gar ford 2
mazda 1
dtype: int64
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
edited Jan 2 at 22:33
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
answered Jan 1 at 14:17
Mykola ZotkoMykola Zotko
1,539416
1,539416
add a comment |
add a comment |
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.
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%2f53991537%2fhow-to-count-a-value-with-json-based-on-another-value-in-the-json-file%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
1
Is Jason the guy who owns the garage?
– wim
Dec 31 '18 at 21:32
@wim I think it's supposed to be JSON. Fixed that.
– Jack Moody
Dec 31 '18 at 21:37
@wim opppsss...
– Shelly875
Dec 31 '18 at 21:38
@Shelly875 I think you will find this answer helpful. Let us know if that solves your problem. It isn't an exact duplicate, but it is pretty closely related.
– Jack Moody
Dec 31 '18 at 21:39