How to make pandas DataFrame from list of dictionaries-CodePudding

So say I have an input which is a list of dictionaries:

in: List[Dict[Decimal, Decimal]] = [{0: 0.05, 0.5: 0.15, 1: 0.1}, {0: 0.06, 0.5: 0.14, 1: 0.2}, {0: 0.09, 0.5: 0.25, 1: 0.17}]

Each dictionary is basically {x_1 : f(x_1), x_2: f(x_2),...}. There are m key-value pairs in a given dictionary, m is fixed.

I want my output to look like

# Index | Dict 1 | Dict 2 | Dict 3
# 0        0.05    0.06    0.09
# 0.5      0.15    0.14    0.25
# 1        0.1     0.2     0.17

So generalised it would be

Index |  Col 1  |  Col 2  | ... | Col n
x_1   | f_1(x_1)| f_2(x_1)| ... | f_n(x_1)
x_2   | f_1(x_2)| f_2(x_2)| ... | f_n(x_2)
...      ...        ...     ...   ...
x_m   | f_1(x_m)| f_2(x_m)| ... | f_n(x_m)

For a fixed and constant n,m where n=len(in) and m= len(in.keys()). The only thing I could think from is to use pd.DataFrame.from_dict() on each individual dict and then pd.concat all of them. I was wondering if there is a more straightforward (and/or fast) way.

CodePudding user response：

See if this helps

import pandas as pd
l=[{0: 0.05, 0.5: 0.15, 1: 0.1}, {0: 0.06, 0.5: 0.14, 1: 0.2}, {0: 0.09, 0.5: 0.25, 1: 0.17}]
df=pd.DataFrame(l)
cols=df.columns
df.set_index(cols,inplace=True)
print(df.head())

You can then transform the cols and index however you like.

CodePudding user response：

You can do:

dict_list = [{0: 0.05, 0.5: 0.15, 1: 0.1}, {0: 0.06, 0.5: 0.14, 1: 0.2}, {0: 0.09, 0.5: 0.25, 1: 0.17}]
df = pd.DataFrame(dict_list).T
df.columns = (df.columns   1).map('Dict{}'.format)

print(df):

     Dict1  Dict2  Dict3
0.0   0.05   0.06   0.09
0.5   0.15   0.14   0.25
1.0   0.10   0.20   0.17

If you want index to be a separate column you can do:

df = df.reset_index().rename(columns={'index':'Index'})