将单个字符串连接到列表中包含的许多其他元素的最干净的方法

Question

What I have:

string = "string"
range_list = list(range(10))

What I want:

['string0',
 'string1',
 'string2',
 'string3',
 'string4',
 'string5',
 'string6',
 'string7',
 'string8',
 'string9']

What I usually do:

import pandas as pd
(string+pd.Series(range_list).astype(str)).tolist()

What I would like to do:
obtain the same expected output from the same input, without importing libraries nor using loops

Since there is probably no way to do this complying my requests, any other solution cleaner and/or more performing than mine is well accepted. Thanks in advice.

Answer 1

You can do this using list comprehension and f-string.

[f"{string}{idx}" for idx in range_list]

Answer 2

You can use map with a function or a lambda to avoid using a loop.

def get_string(x):
    return f'string{x}'

list(map(get_string, range(10)))

or with a lambda:

list(map(lambda x: f'string{x}', range(10)))

For your case, you could write:

list(map(lambda x: f'{string}{x}', range_list))

Answer 3

Since, you want solutions without loops:

string = "string"
range_list = list(range(10))

list(map(lambda x: string + x, np.array(range_list).astype(str)))

Or

list(map(lambda x: 'string{}'.format(x), range_list))

Answer 4

This works too

string = "string"

str_list = [string + str(i) for i in range(10)]