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Create a list with string + range

I need your help:
I want to create a list looking like this ['Unnamed: 16', 'Unnamed: 17', 'Unnamed:18'] for a range (16,60). How can I proceed?
I don't know if my question is clear but it's like doing list(range(16, 60) but with a string before each numbers.
Thank you very much for your help!!

You can use f-strings to do so :
my_list = [f"Unnamed: {i}" for i in range(16, 60)]
# Output
['Unnamed: 16', 'Unnamed: 17', 'Unnamed: 18', 'Unnamed: 19', ...]

I would do it following way
prefix = "Unnamed: "
lst = [prefix + str(i) for i in range(16,25)]
print(lst)
output
['Unnamed: 16', 'Unnamed: 17', 'Unnamed: 18', 'Unnamed: 19', 'Unnamed: 20', 'Unnamed: 21', 'Unnamed: 22', 'Unnamed: 23', 'Unnamed: 24']
Note: I used othre range for brevity sake. You might elect to use one of string formatting method instead.

You can do it using map as,
list(map(lambda x: f'Unnamed: {x}', range(16, 60)))

You can use f strings
name = 'Unamed:'
list = [f"{prefix} {i}" for i in range(16, 60)]
print(list)

my_list = []
for i in range(16, 60):
my_list.append("Unnamed: " + str(i))
print(my_list)

Python, sort with key

I am having trouble understanding key parameter in sorted function in Python.
Let us say, the following list is given: sample_list = ['Date ', 'of', 'birth', 'year 1990', 'month 10', 'day 15'] and I would like to sort only the strings that contain number.
Expected output: ['Date ', 'of', 'birth', 'month 10', 'day 15', 'year 1990']
Until now, I only managed to print the string with the number
def sort_with_key(element_of_list):
if re.search('\d+', element_of_list):
print(element_of_list)
return element_of_list
sorted(sample_list, key = sort_with_key)
But how do I actually sort those elements?
Thank you!

We can try sorting with a lambda:
sample_list = ['Date ', 'of', 'birth', 'year 1990', 'month 10', 'day 15']
sample_list = sorted(sample_list, key=lambda x: int(re.findall(r'\d+', x)[0]) if re.search(r'\d+', x) else 0)
print(sample_list)
This prints:
['Date ', 'of', 'birth', 'month 10', 'day 15', 'year 1990']
The logic used in the lambda is to sort by the number in each list entry, if the entry has a number. Otherwise, it assigns a value of zero to other entries, placing them first in the sort.

If I understand correctly, you want strings with a number to be sorted with this number as key, and strings without a number to be at the beginning?
You need a key that extracts the number from the string. We can use str.isdigit() to extract digits from a string, ''.join() to put these digits back together, and int() to convert to an integer. If there are no digits in the string, we'll return -1 instead, so it comes before all nonnegative numbers.
sample_list = ['Date ', 'of', 'birth', 'year 1990', 'month 10', 'day 15', 'answer 42', 'small number 0', 'large number 8676965', 'no number here']
sample_list.sort(key=lambda s: int(''.join(c for c in s if c.isdigit()) or -1))
print(sample_list)
# ['Date ', 'of', 'birth', 'no number here', 'small number 0', 'month 10', 'day 15', 'answer 42', 'year 1990', 'large number 8676965']

Count how many times a sublist is in a list

I want to know how many times a sublist is in a list next to eachother.
From another question, I received the following code to determinate if a sublist is in a list:
list_sequence = ['Example 64', 'Example 32', 'Example 16']
my_list = ['Example 128', 'Example 64', 'Example 32', 'Example 16', 'Example 256', 'Example 512', 'Example 1024']
print(str(list_sequence)[1:-1] in str(my_list))
But I want to know, how many times the list_sequence is NEXT to each other to determinate a combo. In the top example its 1, but if I would append list_sequence at the start and at the very end, it still would 1. If I would add it right after Example 16 in my_list, it would be 2.

You can use zip and enumerate within a list comprehension to find the respective indices of places where your sub_list occurs. Then select those who's subtraction is equal to 3 (length of sub_list):
In [6]: list_sequence = ['Example 64', 'Example 32', 'Example 16']
...:
...: my_list = ['Example 128', 'Example 64', 'Example 32', 'Example 16', 'Example 64', 'Example 32', 'Example 16', 'Example 256', 'Example 512', 'E
...: xample 1024','Example 64', 'Example 32', 'Example 16']
...:
In [7]: indices = [index for index, (i, j, k) in enumerate(zip(my_list, my_list[1:], my_list[2:])) if list_sequence == [i, j, k]]
Out[7]: [1, 4, 10]
In [8]: sum(2 * (j-i == len(list_sequence)) for i, j in zip(indices, indices[1:]))
Out[8]: 2
You an only use a generator expression within sum to find the number of occurrences:
In [4]: sum(list_sequence == [i, j, k] for i, j, k in zip(my_list, my_list[1:], my_list[2:]))
Out[4]: 1
But note that this will also include overlaps because zip(my_list, my_list[1:], my_list[2:])) will give you all consequent triples.

Count By creating dictionary of list_sequence
list_sequence = ['Example 64', 'Example 32', 'Example 16']
my_list = ['Example 128', 'Example 64', 'Example 32', 'Example 16','Example 16', 'Example 256', 'Example 512', 'Example 1024']
dic=dict()
for i in list_sequence :
for j in my_list :
if i==j:
dic[i]=dic.get(i,0)+1
print(dic)
{'Example 64': 1, 'Example 32': 1, 'Example 16': 2}

How to print part of item in a list? [closed]

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I have a list that looks like this:
['Blake 4', 'Bolt 1', 'De Grasse 3', 'Gatlin 2', 'Simbine 5', 'Youssef Meite 6']
I am trying to print this:
Blake
Bolt
De Grasse
Gatlin
Simbine
Youssef Meite
How do I go about writing a list comprehension that handles this scenario? I tried using split and indexing but nothing I have used has worked.

Assuming all the values keep that pattern, split and join back, ignoring last value in splitted array:
l = ['Blake 4', 'Bolt 1', 'De Grasse 3', 'Gatlin 2', 'Simbine 5', 'Youssef Meite 6']
for x in l:
print(' '.join(x.split()[:-1]))
Otherwise, use regex to eliminate numerals:
import re
l = ['Blake 4', 'Bolt 1', 'De Grasse 3', 'Gatlin 2', 'Simbine 5', 'Youssef Meite 6']
for x in l:
print(re.sub(' \d+', '', x))

list comprehension is useless to print stuff (or any operations where you don't need the return value).
In your case, you could use str.rpartition in a loop to print only the left hand of the rightmost space found in the string:
l =['Blake 4', 'Bolt 1', 'De Grasse 3', 'Gatlin 2', 'Simbine 5', 'Youssef Meite 6']
for s in l:
print(s.rpartition(" ")[0])
Just stripping the last digit off the list would be a good usage of listcomp:
newl = [s.rpartition(" ")[0] for s in l]

Try this:
l = ['Blake 4', 'Bolt 1', 'De Grasse 3', 'Gatlin 2', 'Simbine 5', 'Youssef Meite 6']
for i in l:
print(i[:-2])
Indexing is sufficient for solving your problem.
Based on #Jean-François 's comment, if you are trying to remove all the characters before the last space, you can do this instead:
l = ['Blake 4', 'Bolt 1', 'De Grasse 3', 'Gatlin 2', 'Simbine 5', 'Youssef Meite 6']
for i in l:
print(i[:-i[::-1].index(' ')-1])

Swap an item in list of list in python?

Given the list puzzle above, I need to be able to swap the position of X to left, right, top or bottom. The question is, how do I swap it with the item beside or in the other list in the list puzzle?
puzzle = [[' 1', ' 2', ' 3', ' 4'],
[' 5', ' 6', ' 7', ' 8'],
[' 9', '10', '11', '12'],
['13', '14', '15', ' X']]

#generic swap puzzle[a][b] with puzzle[c][d]
puzzle[a][b], puzzle[c][d] = puzzle[c][d], puzzle[a][b]