如何在列表列表中按空格和引号拆分元素？

Question

I have a list of data df1 , that consists of three lists strings:

df1 = [
    ['1 "P040" 68.13 "P040_1" 2.55 8'],
    ['2 "P040" 46.82 "P040_2" 2.53 8'],
    ['3 "P040" 46.82 "P040_3" 2.51 8']
]

I want to convert it to the following list of lists df2 , without the double quotation marks ( " ):

df2 = [
    ['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
    ['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
    ['3', 'P040', '46.82', 'P040_3', '2.51', '8']
]

I tried the following but does not work well

for row in df1:
    for elem in row:
        elem.strip().split('"')
        elem.strip().split('"')

Answer 1

Here is a simple way to do this, you can replace unwanted quotes and then split by a space to get a list as result using a list comprehension:

df2 = [''.join(row).replace('"', '').split(" ") for row in df1]

print(df2)

Output:

[['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
['3', 'P040', '46.82', 'P040_3', '2.51', '8']]

Answer 2

df2 = []
for row in df1:
    for elem in row:
        df2.append(elem.replace('"', '').split(' '))

Answer 3

Since every list in the list consist of only one element, you don't really have to run through 2 for loops. This can be solved with an one-liner list comprehension:

df1 = [
['1 "P040" 68.13 "P040_1" 2.55 8'],
['2 "P040" 46.82 "P040_2" 2.53 8'],
['3 "P040" 46.82 "P040_3" 2.51 8']
]

df2 = [row[0].replace('"', '').split(' ') for row in df1]

print(df2)

>>> [['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
     ['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
     ['3', 'P040', '46.82', 'P040_3', '2.51', '8']]

Answer 4

You can achieve it with nested list comprehension:

df2 = [[item.strip('""') for item in elem.split()] for row in df1 for elem in row]

Or nested loop with list comprehension:

df2 = []
for row in df1:
    for elem in row:
        df2.append([item.strip('""') for item in elem.split()])

Output:

['1', 'P040', '68.13', 'P040_1', '2.55', '8']
['2', 'P040', '46.82', 'P040_2', '2.53', '8']
['3', 'P040', '46.82', 'P040_3', '2.51', '8']

Answer 5

df1 = [['1 "P040" 68.13 "P040_1" 2.55 8'],
       ['2 "P040" 46.82 "P040_2" 2.53 8'],
       ['3 "P040" 46.82 "P040_3" 2.51 8']]

df1 = [[v.replace(' ','').split('"') for v in l] for l in df1]
print(df1)

Answer 6

You can use a combination of the split() and strip() functions to split on both quotes and spaces.

This example will split the elements in df1 and create a new df2 list:

df2 = []
for row in df1:
    new_row = []
    for elem in row[0].split('"'):
        new_row.extend(elem.strip().split())
    df2.append(new_row)

print(df2)

Answer 7

You can use shlex.split() to split on 'words', where a word might be a quoted string, as follows:

import shlex

for i in range(len(df1)):
    df1[i] = shlex.split(df1[i][0])

This is assuming that each of your list items is always one list containing the string. This modifies df1 in place, to create a new 'df2', it would be:

import shlex

df2 = []
for row in df1:
    df2.append(shlex.split(df1[i][0]))

The output will be:

[
  ['1', 'P040', '68.13', 'P040_1', '2.55', '8'],
  ['2', 'P040', '46.82', 'P040_2', '2.53', '8'],
  ['3', 'P040', '46.82', 'P040_3', '2.51', '8']
]

The advantage of using shlex.split() is that the quoted strings can contain spaces without creating a problem that a plain 'split()' won't solve.