Pandas 子字符串 DataFrame 列

Question

I have a pandas DataFrame, with a column called positions , that includes string values with the syntax of the following examples:

[{'y': 49, 'x': 44}, {'y': 78, 'x': 31}]
[{'y': 1, 'x': 63}, {'y': 0, 'x': 23}]
[{'y': 54, 'x': 9}, {'y': 78, 'x': 3}]

I want to create four new columns in my pandas DataFrame, y_start , x_start , y_end , x_end , that are extractions of only the numbers.

Eg for the example of the first row, my new columns would have the following values:

y_start = 49
x_start = 44
y_end = 78
x_end = 31

To summarise, I am looking to extract just the first, second, third, and four occurrence of numbers and save these to individual columns.

Answer 1

• The first issue is to convert the strings back to dicts, which can be done with ast.literal_eval
• Separate the lists to separate columns with the pandas.DataFrame constructor, because it's faster than using .apply(pd.Series)
  • Pandas split column of lists into multiple columns
• Convert the dicts in each column to separate columns per key, using pandas.json_normalize , .rename the columns, and .concat them together.
• Splitting dictionary/list inside a Pandas Column into Separate Columns doesn't quite answer the question, but it's similar.
• If the data is being loaded from a csv, use the converters parameter with .read_csv .
  • df = pd.read_csv('data.csv', converters={'str_column': literal_eval})

import pandas as pd
from ast import literal_eval

# dataframe
data = {'data': ["[{'y': 49, 'x': 44}, {'y': 78, 'x': 31}]", "[{'y': 1, 'x': 63}, {'y': 0, 'x': 23}]", "[{'y': 54, 'x': 9}, {'y': 78, 'x': 3}]"]}

df = pd.DataFrame(data)

# convert the strings in the data column to dicts
df.data = df.data.apply(literal_eval)

# separate the strings into separate columns
df[['start', 'end']] = pd.DataFrame(df.data.tolist(), index=df.index)

# use json_normalize to convert the dicts to separate columns and join the dataframes with concat
cleaned = pd.concat([pd.json_normalize(df.start).rename(lambda x: f'{x}_start', axis=1), pd.json_normalize(df.end).rename(lambda x: f'{x}_end', axis=1)], axis=1)

# display(cleaned)
   y_start  x_start  y_end  x_end
0       49       44     78     31
1        1       63      0     23
2       54        9     78      3

Answer 2

Convert string to object:

import ast
df['positions'] = df['positions'].apply(ast.literal_eval)

This is one way:

df1 = pd.DataFrame.from_records(pd.DataFrame.from_records(df.positions)[0]).rename(columns={"x":"x_start", "y":"y_start"})    
df2 = pd.DataFrame.from_records(pd.DataFrame.from_records(df.positions)[1]).rename(columns={"x":"x_end", "y":"y_end"})
df_new = pd.concat([df1, df2], axis=1)

another, a little more concise:

df1 = pd.DataFrame(df.positions.to_list())[0].apply(pd.Series).rename(columns={"x":"x_start", "y":"y_start"})
df2 = pd.DataFrame(df.positions.to_list())[1].apply(pd.Series).rename(columns={"x":"x_end", "y":"y_end"})
df_new = pd.concat([df1, df2], axis=1)

I don't know offhand the time or memory performance of how these methods compare.

output (either method):

   y_start  x_start  y_end  x_end
0       49       44     78     31
1        1       63      0     23
2       54        9     78      3

Answer 3

Not so clean but the working way is to write a custom function and apply lambda assuming that all your rows follow the same pattern as provided in your question:

### custom function
def startEndxy(x):
    x = x.split(':')
    return x[1].split(',')[0].replace(' ', ''), x[2].split('},')[0].replace(' ', ''), x[3].split(',')[0].replace(' ', ''), x[4].split('}')[0].replace(' ', '')


### columns creations
df['y_start'] = df['positions'].apply(lambda x: startEndxy(x)[0])
df['x_start'] = df['positions'].apply(lambda x: startEndxy(x)[1])
df['y_end'] = df['positions'].apply(lambda x: startEndxy(x)[2])
df['x_end'] = df['positions'].apply(lambda x: startEndxy(x)[3])

It should give you this output: Output

Answer 4

First reconstruct your series

df = pd.DataFrame(df['position'].tolist()).rename(columns={0: 'starts', 1:'ends'})

              starts               ends
0  {'y': 54, 'x': 9}  {'y': 78, 'x': 3}
1  {'y': 1, 'x': 63}  {'y': 0, 'x': 23}
2  {'y': 54, 'x': 9}  {'y': 78, 'x': 3}

Then assign the start and end columns

starts = pd.DataFrame(df['starts'].tolist()).rename(columns={'y': 'y_start', 'x': 'x_start'})
ends = pd.DataFrame(df['end'].tolist()).rename(columns={'y': 'y_start', 'x': 'x_start'})

df = pd.concat([starts, ends], axis=1)

   y_start  x_start  y_end  x_end
0       54        9     78      3
1        1       63      0     23
2       54        9     78      3