creating dataframe from dictionary of lists

Topics

• pandas

We can create a DataFrame object from a dictionary of lists. The key is the column name.

When you are trying to specify an index for each column value, only the rows with the same index value will be joined. Otherwise, a new row is created, and its columns are filled by NaN if the type is int or float.

The example code below shows both single and multi-level indexing in the DataFrame.

import pandas as pd
 
# example 1: init a dataframe by dict without index
d = {"a": [1, 2, 3, 4], "b": [2, 4, 6, 8]}
df = pd.DataFrame(d)
print(df)
print("---------------------")
 
# example 2: init a dataframe by dict with different index
d = {"a": {"a1":1, "a2":2, "c":3}, "b":{"b1":2, "b2":4, "c":9}}
df = pd.DataFrame(d)
print(df)

   a  b
0  1  2
1  2  4
2  3  6
3  4  8
---------------------
      a    b
a1  1.0  NaN
a2  2.0  NaN
b1  NaN  2.0
b2  NaN  4.0
c   3.0  9.0

Actually, DataFrame also has a from_dict function, which takes a dict of dicts or a dict of array-like sequences and returns a DataFrame.

This function provides a parameter orient with the default value columns. You can assign it with the value index to make the dict keys row labels.

import pandas as pd
 
d = {"a": [1, 2, 3, 4], "b": [2, 4, 6, 8]}
df = pd.DataFrame.from_dict(d)
print("dataframe created from from_dict")
print(df)
print("--------------------")
 
df = pd.DataFrame.from_dict(d, orient="index")
print("dataframe created from from_dict and set the orient")
print(df)

dataframe created from from_dict
   a  b
0  1  2
1  2  4
2  3  6
3  4  8
--------------------
dataframe created from from_dict and set the orient
   0  1  2  3
a  1  2  3  4
b  2  4  6  8

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