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Printing the number of different numbers in python

I would like to ask a question please regarding printing the number of different numbers in python.
for example:
Let us say that I have the following list:
X = [5, 5, 5]
Since here we have only one number, I want to build a code that can recognize that we have only one number here so the output must be:
1
The number is: 5
Let us say that I have the following list:
X = [5,4,5]
Since here we have two numbers (5 and 4), I want to the code to recognize that we have only two numbers here so the output must be:
2
The numbers are: 4, 5
Let us say that I have the following list:
X = [24,24,24,24,24,24,24,24,26,26,26,26,26,26,26,26]
Since here we have two numbers (24 and 26), I want to the code to recognize that we have only two numbers here so the output must be:
2
The numbers are: 24, 26

You could keep track of unique numbers with a set object:
X = [1,2,3,3,3]
S = set(X)
n = len(S)
print(n, S) # 3 {1,2,3}
Bear in mind sets are unordered, so you would need to convert back to a list and sort them if needed.

you can change this list into set, it will remove duplicate, then you can change it again into list.
list(set(X))

You can try numpy.unique, and use len() on the result

May I ask you please if we can use set() to read the data in a specific column in pandas?
For example, I have the following the DataFrame:
df1= [ 0 -10 2 5
1 24 5 10
2 30 3 6
3 30 2 1
4 30 4 5 ]
where the first column is the index..
I tried first to isolate the second column
[-10
24
30
30
30]
using the following: x = pd.DataFrame(df1, coulmn=[0])
Then, I transposed the column using the following XX = x.T
Then, I used set() function.
However, instead of obtaining
[-10 24 30]
I got the following [0 1 2 3 4]
So set() read the index instead of reading the first column

how to write a list in a file with a specific format?

I have a Python list and wanna reprint that in a special way.
input:
trend_end= ['skill1',10,0,13,'skill2',6,1,0,'skill3',5,8,9,'skill4',9,0,1]
I want to write a file like this:
output:
1 2 3
1 10 0 13
2 6 1 0
3 5 8 9
4 9 0 1
Basically, I need to do the following steps:
Separate elements of the list for each skill.
Write them in a table shape, add indices of columns and rows.
I wanna use it as an input of another software. That's why I wanna write a file.
I did this but I know it is wrong, can you see how I can fix it?
f1 = open("data.txt", "a")
for j in trend_end:
f1.write(str(j))
for i in range(1,int(len(trend_end)/df1ana.shape[0])):
G=[trend_end[i*(df1ana.shape[0]-10)- (df1ana.shape[0]-10):i*(df1ana.shape[0]-10)]]
for h in G:
f1.write(i)
f1.write(h)
f1.write('\n')
f.close()
df1ana.shape[0] is 3 in the above example. It is basically the length of data for each skill

Another option that you can try via pandas:
import pandas as pd
pd.DataFrame([trend_end[i+1:i+4] for i in range(0,len(trend_end),4)]).to_csv('data.txt', sep='\t')
OUTPUT:
0 1 2
0 10 0 13
1 6 1 0
2 5 8 9
3 9 0 1

You should iterate over the list in steps of 4, i.e. df1ana.shape[0]+1
steps = df1ana.shape[0]+1
with open("data.txt", "a") as f:
f.write(' ' + ' '.join(range(1, steps)) + '\n') # write header line
for i in range(1, len(trend_end), steps):
f.write(f"{i:<3}")
for j in range(i, i+steps-1):
f.write("f{trend_end[j]:<3}")
f.write("\n")
The :<3 formatting puts each value in a 3-character, left-aligned field.

This should work regardless of the number of groups or the number of records per group. It uses the difference in the size of the full list compared to the integer only list to calculate the number of rows you should have, and uses the ratio of the number of integers over the number of rows to get the number of columns.
import numpy as np
import pandas as pd
digits = [x for x in trend if isinstance(x,int)]
pd.DataFrame(np.reshape(digits,
(int(len(trend)-len(digits)),
int(len(digits)/(len(trend)-len(digits)))))).to_csv('output.csv')

what's the best way to increment a list of numbers

I have a list like this x=[1,2,2,3,1,2,1,1,2,2] where the number is a positive integer that increments by 0 or 1 and sometimes resets to 1, and need to transform it to [1,2,2,3,4,5,6,7,8,8] in an incremental way, where each 1 should be the previous number plus 1 and whatever follows 1 increment accordingly. Is there a simple way to do this via a numpy array etc? I tried using loops but I guess there's a simpler way.

You can use np.add.accumulate():
import numpy as np
x = np.array([1,2,2,3,1,2,1,1,2,2])
x[1:] += np.add.accumulate(x[:-1]*(x[1:]==1))
print(x)
[1 2 2 3 4 5 6 7 8 8]

Slicing large lists based on input

If I have multiple lists such that
hello = [1,3,5,7,9,11,13]
bye = [2,4,6,8,10,12,14]
and the user inputs 3
is there a way to get the output to go back 3 indexes in the list and start there to get:
9 10
11 12
13 14
with tabs \t between each space.
if the user would input 5
the expected output would be
5 6
7 8
9 10
11 12
13 14
I've tried
for i in range(user_input):
print(hello[-i-1], '\t', bye[-i-1])

Just use negative indexies that start from the end minus the user input (-user_input) and move to the the end (-1), something like:
for i in range(-user_input, 0):
print(hello[i], bye[i])

Another zip solution, but one-lined:
for h, b in zip(hello[-user_input:], bye[-user_input:]):
print(h, b, sep='\t')
Avoids converting the result of zip to a list, so the only temporaries are the slices of hello and bye. While iterating by index can avoid those temporaries, in practice it's almost always cleaner and faster to do the slice and iterate the values, as repeated indexing is both unpythonic and surprisingly slow in CPython.

Use negative indexing in the slice.
hello = [1,3,5,7,9,11,13]
print(hello[-3:])
print(hello[-3:-2])
output
[9, 11, 13]
[9]

You can zip the two lists and use itertools.islice to obtain the desired portion of the output:
from itertools import islice
print('\n'.join(map(' '.join, islice(zip(map(str, hello), map(str, bye)), len(hello) - int(input()), len(hello)))))
Given an input of 3, this outputs:
5 6
7 8
9 10
11 12
13 14

You can use zip to return a lists of tuple where the i-th element comes from the i-th iterable argument.
zip_ = list(zip(hello, bye))
for item in zip_[-user_input:]:
print(item[0], '\t' ,item[1])
then use negative index to get what you want.

If you want to analyze the data
I think using pandas.datafrme may be helpful.
INPUT_INDEX = int(input('index='))
df = pd.DataFrame([hello, bye])
df = df.iloc[:, len(df.columns)-INPUT_INDEX:]
for col in df.columns:
h_value, b_value = df[col].values
print(h_value, b_value)
console
index=3
9 10
11 12
13 14

counting T/F values for several conditions

I am a beginner using pandas.
I'm looking for mutations on several patients. I have 16 different conditions. I simply write a code about it but how can do this by for loop? I try to find the changes on MUT column and set them as True and False. Then try to count the True/False numbers. I have done for only 4.
Can you suggest a more simple way, instead of writing the same code 16 times?
s1=df["MUT"]
A_T= s1.str.contains("A:T")
ATnum= A_T.value_counts(sort=True)
s2=df["MUT"]
A_G=s2.str.contains("A:G")
AGnum=A_G.value_counts(sort=True)
s3=df["MUT"]
A_C=s3.str.contains("A:C")
ACnum=A_C.value_counts(sort=True)
s4=df["MUT"]
A__=s4.str.contains("A:-")
A_num=A__.value_counts(sort=True)

I'm not an expert with using Pandas, so don't know if there's a cleaner way of doing this, but perhaps the following might work?
chars = 'TGC-'
nums = {}
for char in chars:
s = df["MUT"]
A = s.str.contains("A:" + char)
num = A.value_counts(sort=True)
nums[char] = num
ATnum = nums['T']
AGnum = nums['G']
# ...etc
Basically, go through each unique character (T, G, C, -) then pull out the values that you need, then finally stick the numbers in a dictionary. Then, once the loop is finished, you can fetch whatever numbers you need back out of the dictionary.

Just use value_counts, this will give you a count of all unique values in your column, no need to create 16 variables:
In [5]:
df = pd.DataFrame({'MUT':np.random.randint(0,16,100)})
df['MUT'].value_counts()
Out[5]:
6 11
14 10
13 9
12 9
1 8
9 7
15 6
11 6
8 5
5 5
3 5
2 5
10 4
4 4
7 3
0 3
dtype: int64