Table with dictionary rewrite into separate columns

[Mystic]

I have a column:

all_rel
{'Иван': 0.358, 'Михаил':0.25, 'Илья': 0.456}
{'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426}
{'Иван': 0.568, 'Михаил':0.145, 'Илья': 0.464}
{'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426}

We need to rewrite the data into several columns in a way:

all_rel	Ivan	Mikhail	Ilya
{'Иван': 0.358, 'Михаил':0.25, 'Илья': 0.456}	0.358	0.25	0.456
{'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426}	0.698	0.125	0.426
{'Иван': 0.568, 'Михаил':0.145, 'Илья': 0.464}	0.568	0.145	0.464
{'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426}	0.698	0.125	0.426

I don't know the exact meaning of the keys of the dictionary and their number, but there's a dictionary with all the possible names. name_dict = {'Иван': 0, 'Михаил':0, 'Илья': 0,'Алексей': 0, 'Андрей': 0}

My code:

for i in range(srt.shape[0]):
    name_dict.update(ast.literal_eval(srt.all_rel[i]))
    srt.rel_dict[i] = name_dict.copy()
for key in srt.rel_dict[i]:
    srt[key] = ' '

for i in range(srt.shape[0]):

for key in srt.rel_dict[i]:

srt[key][i] = srt.rel_dict[i][key]

srt - the name of the date.

The final column may contain columns with other names of the reference dictionary and 0 in each line.

Is there any way to do this differently?

[Burma]

In English SO, there's interesting https://stackoverflow.com/a/55355928/8324991 In your case:

import pandas as pd
df = pd.DataFrame({'all_rel':[

{'Иван': 0.358, 'Михаил':0.25, 'Илья': 0.456},

{'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426},

{'Иван': 0.568, 'Михаил':0.145, 'Илья': 0.464},

{'Иван': 0.698, 'Михаил':0.125, 'Илья': 0.426},

]})
pd.json_normalize(df['all_rel'])

Conclusion:

    Иван    Михаил  Илья

0   0.358   0.250   0.456

1   0.698   0.125   0.426

2   0.568   0.145   0.464

3   0.698   0.125   0.426

There are other interesting options on the reference, but it's kind of fast and simple.

The conversion data with the main date may be combined by index:

df = df.drop(columns='all_rel') 

.join(pd.json_normalize(df['all_rel']))

Or so, just as the respected MaxU says:

df = df.join(pd.json_normalize(df.pop('all_rel')))