In this article we will discuss different techniques to create a DataFrame object from dictionary.
Create DataFrame from Dictionary using default Constructor
DataFrame constructor accepts a data object that can be ndarray, dictionary etc. i.e.
pandas.DataFrame(data=None, index=None, columns=None, dtype=None, copy=False)
But if we are passing a dictionary in data, then it should contain a list like objects in value field like Series, arrays or lists etc i.e.
# Dictionary with list object in values
studentData = {
'name' : ['jack', 'Riti', 'Aadi'],
'age' : [34, 30, 16],
'city' : ['Sydney', 'Delhi', 'New york']
}
On Initialising a DataFrame object with this kind of dictionary, each item (Key / Value pair) in dictionary will be converted to one column i.e. key will become Column Name and list in the value field will be the column data i.e.
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''' Pass dictionary in Dataframe constructor to create a new object keys will be the column names and lists in values will be column data ''' dfObj = pd.DataFrame(studentData)
It will create a DataFrame object like this,
age city name 0 34 Sydney jack 1 30 Delhi Riti 2 16 New york Aadi
All the keys in dictionary will be converted to column names and lists in each its value field will we converted to column Data.
Create DataFrame from Dictionary with custom indexes
We can also pass the index list to the DataFrame constructor to replace the default index list i.e.
# Pass custom names of index as list during initialization dfObj = pd.DataFrame(studentData, index=['a', 'b', 'c'])
It will create a DataFrame object like this,
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age city name a 34 Sydney jack b 30 Delhi Riti c 16 New york Aadi
Create DataFrame from not compatible dictionary
As DataFrame constructor accepts a dictionary which should contain a list like objects in values. But what if we have a dictionary that doesn’t have lists in value i.e.
studentAgeData = {
'Jack' : 12,
'Roma' : 13,
'Ritika' : 10,
'Aadi' : 11
}
If we will directly pass this dictionary to DataFrame constructor then it will throw following error,
ValueError: If using all scalar values, you must pass an index
So, how to create a two column DataFrame object from this kind of dictionary and put all keys and values as these separate columns like this,
0 1 a Roma 13 b Jack 12 c Aadi 11 d Ritika 10
For that we will create a list to tuples (key / value) from this dictionary and pass it to another dataframe constructor that accepts a list i.e.
''' Creating dataframe by converting dict to list of items ''' dfObj = pd.DataFrame(list(studentAgeData.items()), index=['a', 'b', 'c', 'd'])
It will create a DataFrame object like this,
0 1 a Roma 13 b Jack 12 c Aadi 11 d Ritika 10
Create DataFrame from Dictionary and skip data
But we want to create a DataFrame object from dictionary by skipping some of the items. Let’s see how to do that,
Suppose we have dictionary like this,
# Dictionary with list object in values
studentData = {
'name' : ['jack', 'Riti', 'Aadi'],
'age' : [34, 30, 16],
'city' : ['Sydney', 'Delhi', 'New york']
}
Create a DataFrame from this by skipping items with key ‘age’ ,
# Creating Dataframe from Dictionary by Skipping 2nd Item from dict dfObj = pd.DataFrame(studentData, columns=['name', 'city'])
As in columns parameter we provided a list with only two column names. So, DataFrame should contain only 2 columns i.e.
name city 0 jack Sydney 1 Riti Delhi 2 Aadi New york
Create DataFrame from Dictionary with different Orientation
We can create a DataFrame from dictionary using DataFrame.from_dict() function too i.e.
DataFrame.from_dict(data, orient='columns', dtype=None)
It accepts a dictionary and orientation too. By default orientation is columns it means keys in dictionary will be used as columns while creating DataFrame.
We can also pass the orientation as ‘index’, which changes the default orientation and makes the keys in dictionary as index i.e.
Dictionary :
studentData = {
'name' : ['jack', 'Riti', 'Aadi'],
'age' : [34, 30, 16],
'city' : ['Sydney', 'Delhi', 'New york']
}
Create DataFrame with index in orientation i.e.
# Create dataframe from dic and make keys, index in dataframe dfObj = pd.DataFrame.from_dict(studentData, orient='index')
It will create a DataFrame object like this,
0 1 2 name jack Riti Aadi city Sydney Delhi New york age 34 30 16
Create DataFrame from nested Dictionary
Suppose we have a nested dictionary i.e.
# Nested Dictionary
studentData = {
0 : {
'name' : 'Aadi',
'age' : 16,
'city' : 'New york'
},
1 : {
'name' : 'Jack',
'age' : 34,
'city' : 'Sydney'
},
2 : {
'name' : 'Riti',
'age' : 30,
'city' : 'Delhi'
}
}
It has 3 items in it and each item contains a dictionary in value field which internally contains the same keys but with different value.
We can directly pass it in DataFrame constructor, but it will use the keys of dict as columns and DataFrame object like this will be generated i.e.
''' Create dataframe from nested dictionary ''' dfObj = pd.DataFrame(studentData)
It will create a DataFrame object like this,
0 1 2 age 16 34 30 city New york Sydney Delhi name Aadi Jack Riti
Now let’s transpose this matrix to swap the column with indexes i.e. data will be more readable with this i.e.
# Transpose dataframe object dfObj = dfObj.transpose()
Now contents of DataFrame will be like this,
age city name 0 16 New york Aadi 1 34 Sydney Jack 2 30 Delhi Riti
Complete example is as follows,
import pandas as pd
def main():
# Dictionary with list object in values
studentData = {
'name' : ['jack', 'Riti', 'Aadi'],
'age' : [34, 30, 16],
'city' : ['Sydney', 'Delhi', 'New york']
}
print('Creating Dataframe from Dictionary')
'''
Pass dictionary in Dataframe constructor to create a new object
keys will be the column names and lists in values will be column data
'''
dfObj = pd.DataFrame(studentData)
# Print data frame object on console
print(dfObj)
print('Creating Dataframe from Dictionary and Custom Indexes')
# Pass custom names of index as list during initialization
dfObj = pd.DataFrame(studentData, index=['a', 'b', 'c'])
# Print dataframe object on console
print(dfObj)
print('Creating Dataframe from non compatible Dictionary')
studentAgeData = {
'Jack' : 12,
'Roma' : 13,
'Ritika' : 10,
'Aadi' : 11
}
'''
Creating dataframe by converting dict to list of items
'''
dfObj = pd.DataFrame(list(studentAgeData.items()), index=['a', 'b', 'c', 'd'])
# Print Dataframe object on console
print(dfObj)
print('Creating Dataframe from Dictionary by Skipping data')
studentData = {
'name' : ['jack', 'Riti', 'Aadi'],
'age' : [34, 30, 16],
'city' : ['Sydney', 'Delhi', 'New york']
}
# Creating Dataframe from Dictionary by Skipping 2nd Item from dict
dfObj = pd.DataFrame(studentData, columns=['name', 'city'])
# Print Dataframe object on console
print(dfObj)
print('Creating Dataframe from Dictionary With different orientation')
# Create dataframe from dic and make keys, index in dataframe
dfObj = pd.DataFrame.from_dict(studentData, orient='index')
print(dfObj)
print('Creating Dataframe from nested Dictionary')
# Nested Dictionary
studentData = {
0 : {
'name' : 'Aadi',
'age' : 16,
'city' : 'New york'
},
1 : {
'name' : 'Jack',
'age' : 34,
'city' : 'Sydney'
},
2 : {
'name' : 'Riti',
'age' : 30,
'city' : 'Delhi'
}
}
'''
Create dataframe from nested dictionary
'''
dfObj = pd.DataFrame(studentData)
# Print Dataframe object on console
print(dfObj)
print("Transpose the dictionary")
# Transpose dataframe object
dfObj = dfObj.transpose()
print(dfObj)
if __name__ == '__main__':
main()
Output:
Creating Dataframe from Dictionary
age city name
0 34 Sydney jack
1 30 Delhi Riti
2 16 New york Aadi
Creating Dataframe from Dictionary and Custom Indexes
age city name
a 34 Sydney jack
b 30 Delhi Riti
c 16 New york Aadi
Creating Dataframe from non compatible Dictionary
0 1
a Aadi 11
b Roma 13
c Jack 12
d Ritika 10
Creating Dataframe from Dictionary by Skipping data
name city
0 jack Sydney
1 Riti Delhi
2 Aadi New york
Creating Dataframe from Dictionary With different orientation
0 1 2
age 34 30 16
name jack Riti Aadi
city Sydney Delhi New york
Creating Dataframe from nested Dictionary
0 1 2
age 16 34 30
city New york Sydney Delhi
name Aadi Jack Riti
Transpose the dictionary
age city name
0 16 New york Aadi
1 34 Sydney Jack
2 30 Delhi Riti
The list tip and transpose was exactly what I was looking for. Thank you!
dataframe with examples clearly makes concepts easy to understand.