Convert list of dictionaries to a pandas DataFrame
Question:
How can I convert a list of dictionaries into a DataFrame? Given:
[{'points': 50, 'time': '5:00', 'year': 2010},
{'points': 25, 'time': '6:00', 'month': "february"},
{'points':90, 'time': '9:00', 'month': 'january'},
{'points_h1':20, 'month': 'june'}]
I want to turn the above into a DataFrame:
month points points_h1 time year
0 NaN 50 NaN 5:00 2010
1 february 25 NaN 6:00 NaN
2 january 90 NaN 9:00 NaN
3 june NaN 20 NaN NaN
Note: Order of the columns does not matter.
Answers:
If ds is a list of dicts:
df = pd.DataFrame(ds)
Note: this does not work with nested data.
In pandas 16.2, I had to do pd.DataFrame.from_records(d) to get this to work.
You can also use pd.DataFrame.from_dict(d) as :
In [8]: d = [{'points': 50, 'time': '5:00', 'year': 2010},
...: {'points': 25, 'time': '6:00', 'month': "february"},
...: {'points':90, 'time': '9:00', 'month': 'january'},
...: {'points_h1':20, 'month': 'june'}]
In [12]: pd.DataFrame.from_dict(d)
Out[12]:
month points points_h1 time year
0 NaN 50.0 NaN 5:00 2010.0
1 february 25.0 NaN 6:00 NaN
2 january 90.0 NaN 9:00 NaN
3 june NaN 20.0 NaN NaN
How do I convert a list of dictionaries to a pandas DataFrame?
The other answers are correct, but not much has been explained in terms of advantages and limitations of these methods. The aim of this post will be to show examples of these methods under different situations, discuss when to use (and when not to use), and suggest alternatives.
DataFrame(), DataFrame.from_records(), and .from_dict()
Depending on the structure and format of your data, there are situations where either all three methods work, or some work better than others, or some don’t work at all.
Consider a very contrived example.
np.random.seed(0)
data = pd.DataFrame(
np.random.choice(10, (3, 4)), columns=list('ABCD')).to_dict('r')
print(data)
[{'A': 5, 'B': 0, 'C': 3, 'D': 3},
{'A': 7, 'B': 9, 'C': 3, 'D': 5},
{'A': 2, 'B': 4, 'C': 7, 'D': 6}]
This list consists of "records" with every keys present. This is the simplest case you could encounter.
# The following methods all produce the same output.
pd.DataFrame(data)
pd.DataFrame.from_dict(data)
pd.DataFrame.from_records(data)
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
Word on Dictionary Orientations: orient='index'/'columns'
Before continuing, it is important to make the distinction between the different types of dictionary orientations, and support with pandas. There are two primary types: "columns", and "index".
orient='columns'
Dictionaries with the "columns" orientation will have their keys correspond to columns in the equivalent DataFrame.
For example, data above is in the "columns" orient.
data_c = [
{'A': 5, 'B': 0, 'C': 3, 'D': 3},
{'A': 7, 'B': 9, 'C': 3, 'D': 5},
{'A': 2, 'B': 4, 'C': 7, 'D': 6}]
pd.DataFrame.from_dict(data_c, orient='columns')
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
Note: If you are using pd.DataFrame.from_records, the orientation is assumed to be "columns" (you cannot specify otherwise), and the dictionaries will be loaded accordingly.
orient='index'
With this orient, keys are assumed to correspond to index values. This kind of data is best suited for pd.DataFrame.from_dict.
data_i ={
0: {'A': 5, 'B': 0, 'C': 3, 'D': 3},
1: {'A': 7, 'B': 9, 'C': 3, 'D': 5},
2: {'A': 2, 'B': 4, 'C': 7, 'D': 6}}
pd.DataFrame.from_dict(data_i, orient='index')
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
This case is not considered in the OP, but is still useful to know.
Setting Custom Index
If you need a custom index on the resultant DataFrame, you can set it using the index=... argument.
pd.DataFrame(data, index=['a', 'b', 'c'])
# pd.DataFrame.from_records(data, index=['a', 'b', 'c'])
A B C D
a 5 0 3 3
b 7 9 3 5
c 2 4 7 6
This is not supported by pd.DataFrame.from_dict.
Dealing with Missing Keys/Columns
All methods work out-of-the-box when handling dictionaries with missing keys/column values. For example,
data2 = [
{'A': 5, 'C': 3, 'D': 3},
{'A': 7, 'B': 9, 'F': 5},
{'B': 4, 'C': 7, 'E': 6}]
# The methods below all produce the same output.
pd.DataFrame(data2)
pd.DataFrame.from_dict(data2)
pd.DataFrame.from_records(data2)
A B C D E F
0 5.0 NaN 3.0 3.0 NaN NaN
1 7.0 9.0 NaN NaN NaN 5.0
2 NaN 4.0 7.0 NaN 6.0 NaN
Reading Subset of Columns
"What if I don’t want to read in every single column"? You can easily specify this using the columns=... parameter.
For example, from the example dictionary of data2 above, if you wanted to read only columns "A’, ‘D’, and ‘F’, you can do so by passing a list:
pd.DataFrame(data2, columns=['A', 'D', 'F'])
# pd.DataFrame.from_records(data2, columns=['A', 'D', 'F'])
A D F
0 5.0 3.0 NaN
1 7.0 NaN 5.0
2 NaN NaN NaN
This is not supported by pd.DataFrame.from_dict with the default orient "columns".
pd.DataFrame.from_dict(data2, orient='columns', columns=['A', 'B'])
ValueError: cannot use columns parameter with orient='columns'
Reading Subset of Rows
Not supported by any of these methods directly. You will have to iterate over your data and perform a reverse delete in-place as you iterate. For example, to extract only the 0th and 2nd rows from data2 above, you can use:
rows_to_select = {0, 2}
for i in reversed(range(len(data2))):
if i not in rows_to_select:
del data2[i]
pd.DataFrame(data2)
# pd.DataFrame.from_dict(data2)
# pd.DataFrame.from_records(data2)
A B C D E
0 5.0 NaN 3 3.0 NaN
1 NaN 4.0 7 NaN 6.0
The Panacea: json_normalize for Nested Data
A strong, robust alternative to the methods outlined above is the json_normalize function which works with lists of dictionaries (records), and in addition can also handle nested dictionaries.
pd.json_normalize(data)
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
pd.json_normalize(data2)
A B C D E
0 5.0 NaN 3 3.0 NaN
1 NaN 4.0 7 NaN 6.0
Again, keep in mind that the data passed to json_normalize needs to be in the list-of-dictionaries (records) format.
As mentioned, json_normalize can also handle nested dictionaries. Here’s an example taken from the documentation.
data_nested = [
{'counties': [{'name': 'Dade', 'population': 12345},
{'name': 'Broward', 'population': 40000},
{'name': 'Palm Beach', 'population': 60000}],
'info': {'governor': 'Rick Scott'},
'shortname': 'FL',
'state': 'Florida'},
{'counties': [{'name': 'Summit', 'population': 1234},
{'name': 'Cuyahoga', 'population': 1337}],
'info': {'governor': 'John Kasich'},
'shortname': 'OH',
'state': 'Ohio'}
]
pd.json_normalize(data_nested,
record_path='counties',
meta=['state', 'shortname', ['info', 'governor']])
name population state shortname info.governor
0 Dade 12345 Florida FL Rick Scott
1 Broward 40000 Florida FL Rick Scott
2 Palm Beach 60000 Florida FL Rick Scott
3 Summit 1234 Ohio OH John Kasich
4 Cuyahoga 1337 Ohio OH John Kasich
For more information on the meta and record_path arguments, check out the documentation.
Summarising
Here’s a table of all the methods discussed above, along with supported features/functionality.
* Use orient='columns' and then transpose to get the same effect as orient='index'.
The easiest way I have found to do it is like this:
dict_count = len(dict_list)
df = pd.DataFrame(dict_list[0], index=[0])
for i in range(1,dict_count-1):
df = df.append(dict_list[i], ignore_index=True)
Pyhton3:
Most of the solutions listed previously work. However, there are instances when row_number of the dataframe is not required and the each row (record) has to be written individually.
The following method is useful in that case.
import csv
my file= 'C:UsersJohnDesktopexport_dataframe.csv'
records_to_save = data2 #used as in the thread.
colnames = list[records_to_save[0].keys()]
# remember colnames is a list of all keys. All values are written corresponding
# to the keys and "None" is specified in case of missing value
with open(myfile, 'w', newline="",encoding="utf-8") as f:
writer = csv.writer(f)
writer.writerow(colnames)
for d in records_to_save:
writer.writerow([d.get(r, "None") for r in colnames])
I have the following list of dicts with datetime keys and int values:
list = [{datetime.date(2022, 2, 10): 7}, {datetime.date(2022, 2, 11): 1}, {datetime.date(2022, 2, 11): 1}]
I had a problem to convert it to Dataframe with the methods above as it created Dataframe with columns with dates…
My solution:
df = pd.DataFrame()
for i in list:
temp_df = pd.DataFrame.from_dict(i, orient='index')
df = df.append(temp_df)
How can I convert a list of dictionaries into a DataFrame? Given:
[{'points': 50, 'time': '5:00', 'year': 2010},
{'points': 25, 'time': '6:00', 'month': "february"},
{'points':90, 'time': '9:00', 'month': 'january'},
{'points_h1':20, 'month': 'june'}]
I want to turn the above into a DataFrame:
month points points_h1 time year
0 NaN 50 NaN 5:00 2010
1 february 25 NaN 6:00 NaN
2 january 90 NaN 9:00 NaN
3 june NaN 20 NaN NaN
Note: Order of the columns does not matter.
If ds is a list of dicts:
df = pd.DataFrame(ds)
Note: this does not work with nested data.
In pandas 16.2, I had to do pd.DataFrame.from_records(d) to get this to work.
You can also use pd.DataFrame.from_dict(d) as :
In [8]: d = [{'points': 50, 'time': '5:00', 'year': 2010},
...: {'points': 25, 'time': '6:00', 'month': "february"},
...: {'points':90, 'time': '9:00', 'month': 'january'},
...: {'points_h1':20, 'month': 'june'}]
In [12]: pd.DataFrame.from_dict(d)
Out[12]:
month points points_h1 time year
0 NaN 50.0 NaN 5:00 2010.0
1 february 25.0 NaN 6:00 NaN
2 january 90.0 NaN 9:00 NaN
3 june NaN 20.0 NaN NaN
How do I convert a list of dictionaries to a pandas DataFrame?
The other answers are correct, but not much has been explained in terms of advantages and limitations of these methods. The aim of this post will be to show examples of these methods under different situations, discuss when to use (and when not to use), and suggest alternatives.
DataFrame(), DataFrame.from_records(), and .from_dict()
Depending on the structure and format of your data, there are situations where either all three methods work, or some work better than others, or some don’t work at all.
Consider a very contrived example.
np.random.seed(0)
data = pd.DataFrame(
np.random.choice(10, (3, 4)), columns=list('ABCD')).to_dict('r')
print(data)
[{'A': 5, 'B': 0, 'C': 3, 'D': 3},
{'A': 7, 'B': 9, 'C': 3, 'D': 5},
{'A': 2, 'B': 4, 'C': 7, 'D': 6}]
This list consists of "records" with every keys present. This is the simplest case you could encounter.
# The following methods all produce the same output.
pd.DataFrame(data)
pd.DataFrame.from_dict(data)
pd.DataFrame.from_records(data)
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
Word on Dictionary Orientations: orient='index'/'columns'
Before continuing, it is important to make the distinction between the different types of dictionary orientations, and support with pandas. There are two primary types: "columns", and "index".
orient='columns'
Dictionaries with the "columns" orientation will have their keys correspond to columns in the equivalent DataFrame.
For example, data above is in the "columns" orient.
data_c = [
{'A': 5, 'B': 0, 'C': 3, 'D': 3},
{'A': 7, 'B': 9, 'C': 3, 'D': 5},
{'A': 2, 'B': 4, 'C': 7, 'D': 6}]
pd.DataFrame.from_dict(data_c, orient='columns')
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
Note: If you are using pd.DataFrame.from_records, the orientation is assumed to be "columns" (you cannot specify otherwise), and the dictionaries will be loaded accordingly.
orient='index'
With this orient, keys are assumed to correspond to index values. This kind of data is best suited for pd.DataFrame.from_dict.
data_i ={
0: {'A': 5, 'B': 0, 'C': 3, 'D': 3},
1: {'A': 7, 'B': 9, 'C': 3, 'D': 5},
2: {'A': 2, 'B': 4, 'C': 7, 'D': 6}}
pd.DataFrame.from_dict(data_i, orient='index')
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
This case is not considered in the OP, but is still useful to know.
Setting Custom Index
If you need a custom index on the resultant DataFrame, you can set it using the index=... argument.
pd.DataFrame(data, index=['a', 'b', 'c'])
# pd.DataFrame.from_records(data, index=['a', 'b', 'c'])
A B C D
a 5 0 3 3
b 7 9 3 5
c 2 4 7 6
This is not supported by pd.DataFrame.from_dict.
Dealing with Missing Keys/Columns
All methods work out-of-the-box when handling dictionaries with missing keys/column values. For example,
data2 = [
{'A': 5, 'C': 3, 'D': 3},
{'A': 7, 'B': 9, 'F': 5},
{'B': 4, 'C': 7, 'E': 6}]
# The methods below all produce the same output.
pd.DataFrame(data2)
pd.DataFrame.from_dict(data2)
pd.DataFrame.from_records(data2)
A B C D E F
0 5.0 NaN 3.0 3.0 NaN NaN
1 7.0 9.0 NaN NaN NaN 5.0
2 NaN 4.0 7.0 NaN 6.0 NaN
Reading Subset of Columns
"What if I don’t want to read in every single column"? You can easily specify this using the columns=... parameter.
For example, from the example dictionary of data2 above, if you wanted to read only columns "A’, ‘D’, and ‘F’, you can do so by passing a list:
pd.DataFrame(data2, columns=['A', 'D', 'F'])
# pd.DataFrame.from_records(data2, columns=['A', 'D', 'F'])
A D F
0 5.0 3.0 NaN
1 7.0 NaN 5.0
2 NaN NaN NaN
This is not supported by pd.DataFrame.from_dict with the default orient "columns".
pd.DataFrame.from_dict(data2, orient='columns', columns=['A', 'B'])
ValueError: cannot use columns parameter with orient='columns'
Reading Subset of Rows
Not supported by any of these methods directly. You will have to iterate over your data and perform a reverse delete in-place as you iterate. For example, to extract only the 0th and 2nd rows from data2 above, you can use:
rows_to_select = {0, 2}
for i in reversed(range(len(data2))):
if i not in rows_to_select:
del data2[i]
pd.DataFrame(data2)
# pd.DataFrame.from_dict(data2)
# pd.DataFrame.from_records(data2)
A B C D E
0 5.0 NaN 3 3.0 NaN
1 NaN 4.0 7 NaN 6.0
The Panacea: json_normalize for Nested Data
A strong, robust alternative to the methods outlined above is the json_normalize function which works with lists of dictionaries (records), and in addition can also handle nested dictionaries.
pd.json_normalize(data)
A B C D
0 5 0 3 3
1 7 9 3 5
2 2 4 7 6
pd.json_normalize(data2)
A B C D E
0 5.0 NaN 3 3.0 NaN
1 NaN 4.0 7 NaN 6.0
Again, keep in mind that the data passed to json_normalize needs to be in the list-of-dictionaries (records) format.
As mentioned, json_normalize can also handle nested dictionaries. Here’s an example taken from the documentation.
data_nested = [
{'counties': [{'name': 'Dade', 'population': 12345},
{'name': 'Broward', 'population': 40000},
{'name': 'Palm Beach', 'population': 60000}],
'info': {'governor': 'Rick Scott'},
'shortname': 'FL',
'state': 'Florida'},
{'counties': [{'name': 'Summit', 'population': 1234},
{'name': 'Cuyahoga', 'population': 1337}],
'info': {'governor': 'John Kasich'},
'shortname': 'OH',
'state': 'Ohio'}
]
pd.json_normalize(data_nested,
record_path='counties',
meta=['state', 'shortname', ['info', 'governor']])
name population state shortname info.governor
0 Dade 12345 Florida FL Rick Scott
1 Broward 40000 Florida FL Rick Scott
2 Palm Beach 60000 Florida FL Rick Scott
3 Summit 1234 Ohio OH John Kasich
4 Cuyahoga 1337 Ohio OH John Kasich
For more information on the meta and record_path arguments, check out the documentation.
Summarising
Here’s a table of all the methods discussed above, along with supported features/functionality.
* Use orient='columns' and then transpose to get the same effect as orient='index'.
The easiest way I have found to do it is like this:
dict_count = len(dict_list)
df = pd.DataFrame(dict_list[0], index=[0])
for i in range(1,dict_count-1):
df = df.append(dict_list[i], ignore_index=True)
Pyhton3:
Most of the solutions listed previously work. However, there are instances when row_number of the dataframe is not required and the each row (record) has to be written individually.The following method is useful in that case.
import csv
my file= 'C:UsersJohnDesktopexport_dataframe.csv'
records_to_save = data2 #used as in the thread.
colnames = list[records_to_save[0].keys()]
# remember colnames is a list of all keys. All values are written corresponding
# to the keys and "None" is specified in case of missing value
with open(myfile, 'w', newline="",encoding="utf-8") as f:
writer = csv.writer(f)
writer.writerow(colnames)
for d in records_to_save:
writer.writerow([d.get(r, "None") for r in colnames])
I have the following list of dicts with datetime keys and int values:
list = [{datetime.date(2022, 2, 10): 7}, {datetime.date(2022, 2, 11): 1}, {datetime.date(2022, 2, 11): 1}]
I had a problem to convert it to Dataframe with the methods above as it created Dataframe with columns with dates…
My solution:
df = pd.DataFrame()
for i in list:
temp_df = pd.DataFrame.from_dict(i, orient='index')
df = df.append(temp_df)