Find a list in a row

Question

I see many examples of finding strings in strings or searching strings in lists, but how to find a list in a string. For example, I have a csv file with data columns, and the last column is a row, or sometimes it is a list. Here is a subset of the data showing only the last 3 columns.

TRUE, 93877, S26476961
TRUE, 93878, ['S26489167', 'S26492524']
FALSE, 93879, S26476962
FALSE, 93880, ['S26489168', 'S26492527', 'S26492528']

At first I tried to parse every whole line of the csv file with a comma, but it parses the commas in the list (creating additional columns). I just want the list to be recognized as a single piece of data, so I can work with it as a list of n elements. The @TemporalWolf comment helps a lot, because if I use the csv module (specifically csv.reader) as such ...

reader = csv.reader(inFile)
for row in reader:
    print(row)

It stores the list in one column. Now the problem remains that this is just a string. In other words, row[n][0]returns the left bracket ( [), but I want to do this in a list.

Answer 1

It depends on the character 'with which the items in your list are quoted. Using this information, it is separated only by commas, not followed or preceding this character, using a regular expression:

import re
import pandas as pd
import io


text = """TRUE, 93877, S26476961
TRUE, 93878, ['S26489167', 'S26492524']
FALSE, 93879, S26476962
FALSE, 93880, ['S26489168', 'S26492527', 'S26492528']"""

with io.StringIO(text) as f:
    for line in f:
        print(re.split("(?<!'), (?!')", line.strip()))


# ['TRUE', '93877', 'S26476961']
# ['TRUE', '93878', "['S26489167', 'S26492524']"]
# ['FALSE', '93879', 'S26476962']
# ['FALSE', '93880', "['S26489168', 'S26492527', 'S26492528']"]

# Or with pandas

with io.StringIO(text) as f:
    print(pd.read_csv(f,
                  header=None,
                  sep="(?<!'), (?!')",
                  engine='python'))

#        0      1                                        2
# 0   True  93877                                S26476961
# 1   True  93878               ['S26489167', 'S26492524']
# 2  False  93879                                S26476962
# 3  False  93880  ['S26489168', 'S26492527', 'S26492528']

Edit:

If you are using python2, you will need to convert the text to unicode (putting a character uin front of the text) in order to be able to use io.StringIO:

import re
import pandas as pd
import io


text = u"""TRUE, 93877, S26476961
TRUE, 93878, ['S26489167', 'S26492524']
FALSE, 93879, S26476962
FALSE, 93880, ['S26489168', 'S26492527', 'S26492528']"""

with io.StringIO(text) as f:
    for line in f:
        print(re.split("(?<!'), (?!')", line.strip()))


# ['TRUE', '93877', 'S26476961']
# ['TRUE', '93878', "['S26489167', 'S26492524']"]
# ['FALSE', '93879', 'S26476962']
# ['FALSE', '93880', "['S26489168', 'S26492527', 'S26492528']"]

# Or with pandas

with io.StringIO(text) as f:
    print(pd.read_csv(f,
                  header=None,
                  sep="(?<!'), (?!')",
                  engine='python'))

#        0      1                                        2
# 0   True  93877                                S26476961
# 1   True  93878               ['S26489167', 'S26492524']
# 2  False  93879                                S26476962
# 3  False  93880  ['S26489168', 'S26492527', 'S26492528']

Edit 2:

', :

import ast
import re


with io.StringIO(text) as f:
    for line in f:
        parts = re.split(", (?=\[)", line.strip())
        line = []
        for part in parts:
            if all(char in part for char in ('[]')):
                line.append(ast.literal_eval(part))
            else:
                line += part.split(", ")
        print(line)

, , :

• , , , . , list ast.literal_eval .
• .

, .

, .

Answer 2

: Python. , , :

import ast

def get_columns(line):
    def valid(code):
        try:
            ast.parse(code.strip())
        except SyntaxError:
            return False
        return True
    sections = line.split(',')
    columns = []
    for i, section in enumerate(sections):
        if i == len(sections) - 1 or valid(section):
            columns.append(section)
        else:
            sections[i + 1] = ','.join([section, sections[i + 1]])
    return columns

with open(inFile) as f:
    for line in f:
        for column in get_columns(line):
            print(column)

, . , , "" , .

Python 2 3.

Answer 3

This is the opposite approach. It finds lists in the data returned csv.readerby checking the leading [and trailing ]in the elements of the string.

import csv 

def find_lists(row):
    sublist = []
    for item in row:
        if not sublist:
            if item.startswith('['):
                if item.endswith(']'):
                    yield [item[1:-1]]
                else:
                    sublist.append(item[1:])
            else:
                yield item
        else:
            if item.endswith(']'):
                sublist.append(item[:-1])
                yield sublist
                sublist = []
            else:
                sublist.append(item)
    for item in sublist:
        yield item

with open('test.csv') as infile:
    reader = csv.reader(infile, skipinitialspace=True)
    for row in reader:
        print(list(find_lists(row)))