How to Remove Square Brackets from List in Python

How to remove square brackets from list in Python?

You could convert it to a string instead of printing the list directly:

print(", ".join(LIST))

If the elements in the list aren't strings, you can convert them to string using either repr (if you want quotes around strings) or str (if you don't), like so:

LIST = [1, "foo", 3.5, { "hello": "bye" }]
print( ", ".join( repr(e) for e in LIST ) )

Which gives the output:

1, 'foo', 3.5, {'hello': 'bye'}

Remove square brackets from lists in a column

IIUC, try this for string storage or list storge:

# String object
df_string = pd.DataFrame({'Fruits':['[apple, kiwi, pineapple, mango]',
                                     '[strawberry, cherry, kiwi, orange]']})

df_string['Fruits'].str.replace('\[|\]','', regex=True)

Output:

0       apple, kiwi, pineapple, mango
1    strawberry, cherry, kiwi, orange
Name: Fruits, dtype: object

Or,

#List object
df_list = pd.DataFrame({'Fruits':[['apple', 'kiwi', 'pineapple', 'mango'],
                                     ['strawberry', 'cherry', 'kiwi', 'orange']]})

df_list['Fruits'].agg(', '.join)

Output:

0       apple, kiwi, pineapple, mango
1    strawberry, cherry, kiwi, orange
Name: Fruits, dtype: object

Note: df_list and df_string string representation looks identical.

print(df_list)
                               Fruits
0     [apple, kiwi, pineapple, mango]
1  [strawberry, cherry, kiwi, orange]

print(df_string)
                              Fruits
0     [apple, kiwi, pineapple, mango]
1  [strawberry, cherry, kiwi, orange]

how to remove square brackets and quotation mark when convert array to list in python

If you have nested lists [['a'], ['b'], ['c']] then you can use for-loop to make it flatten ['a', 'b', 'c']

data = [ ['a'], ['b'], ['c']]

flatten = [row[0] for row in data]

print(flatten)

Or you can also use fact that ["a"] + ["b"] gives ["a", "b"] - so you can use sum() with [] as starting value

data = [['a'], ['b'], ['c']]

flatten = sum(data, [])

print(flatten)

And if you have numpy.array then you could simply use arr.flatten()

import numpy as np

data = [['a'], ['b'], ['c']]
arr = np.array(data)

flatten = arr.flatten()

print(flatten)

BUT ... images show that you have [['X', 'Y'], ['a'], ['b'], ['c']] and first element has two values - and this need different method to create flatten ['X Y', 'a', 'b', 'c']. It needs to use for-loop with join()

data = [['X', 'Y'], ['a'], ['b'], ['c']]

flatten = [' '.join(row) for row in data]

print(flatten)

The same using map()

data = [['X', 'Y'], ['a'], ['b'], ['c']]

flatten = list(map(",".join, data))

print(flatten)

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And when you have flatten list then your code+

rows_list = [flatten]

df = pd.DataFrame(rows_list)
df = df.T

print(df)

gives

     0
0  X Y
1    a
2    b
3    c

without [] and ''

---

BTW:

If you would creat dictionary rows_list[a] = b (after converting a to string and b to flatten list) then you wouldn't need to transpose df = df.T

import pandas as pd

a = [['X', 'Y']]
b = [['a'], ['b'], ['c']]

print('a:', a)
print('b:', b)
print('---')

a = " ".join(sum(a, []))
b = sum(b, [])

print('a:', a)
print('b:', b)
print('---')

rows = dict()
rows[a] = b

df = pd.DataFrame(rows)
print(df)

gives

a: [['X', 'Y']]
b: [['a'], ['b'], ['c']]
---
a: X Y
b: ['a', 'b', 'c']
---
  X Y
0   a
1   b
2   c

Remove Square Bracket and Breakdown List elements

Two good ways:

out = list(x for y in lst for x in y)
...or...
out = sum(lst,[])

how to remove the square bracket that are leftover in list with python?

You can try this below :

    a = [[[1, 3, 2, 4], [1, 3, 2, 6]],
         [[2, 4], [2, 6]],
         [[3, 2, 4], [3, 2, 6]],
         [[4]],
         [[5, 4]],
         [[6]]]
    output = [elem for output_list in a for elem in output_list]
    print(output)

Remove square brackets from a list of characters

[x for x in a if x not in "[]"]

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