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I want to count the frequency of 4 letters at every position across strings. The letters are A, T, G, C
TGAGGTAGTAGTTTGTGCTGTTAT
TAGTAGTTTGTGCTGTTA
TGAGGTAGTAGTTTGTAC
TGAGAACTGAATTCCATAGG
desired output:
Pos1 Pos2 Pos3 and so on.
A 0 1
T 4 0
C 0 0
G 0 3
So far I have used an R package called Biostrings, which works, but I wonder if perl would do this?
For the record, for
x = "TGAGGTAGTAGTTTGTGCTGTTAT
TAGTAGTTTGTGCTGTTA
TGAGGTAGTAGTTTGTAC
TGAGAACTGAATTCCATAGG"
a Biostrings solution is
library(Biostrings)
consensusMatrix(DNAStringSet(strsplit(x, "\n")[[1]]))
which will be fast for millions of sequences.
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Could anyone please help me in understanding the question for this.
I got this from https://www.geeksforgeeks.org/python-program-for-nth-multiple-of-a-number-in-fibonacci-series/
Given two integers n and k. Find position the n\’th multiple of K in the Fibonacci series.
Examples:
Input : k = 2, n = 3
Output : 9
3'rd multiple of 2 in Fibonacci Series is 34
which appears at position 9.
Input : k = 4, n = 5
Output : 30
5'th multiple of 5 in Fibonacci Series is 832040
which appears at position 30.
You are interested in the numbers that are divisible by k in the fibonacci sequence. Among these numbers, you seek the index of the nth, that is, the index i such that the ith fibonacci number is divisible by k, and such that there are exactly n-1 fibonacci numbers before that one that are also divisible by k.
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I'm new to Python and I wanted to know that how can I subtract All the Elements in A List in Python
For Eg -
list [5,6,7,8,9]
I want to subtract it like this = 5 - 6 - 7 - 8 - 9
Use reduce, which expands out to subtracting all elements.
>>> from functools import reduce
>>> reduce(lambda x, y: x - y, [5,6,7,8,9])
-25
To subtract each number from the first one, you can subtract the sum of them:
>>> mylist = [5,6,7,8,9]
>>> print(mylist[0] - sum(mylist[1:]))
-25
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In the date_died column of the dataset, I want to convert the column into a column with the heading 'Survived' where: if date_died is 9999-99-99 then survived = 1 else survived = 0 [1]. I want to use that column as my output data frame. I am an absolute beginner at ML and python.
You can use np.where for this,
df['Survived'] = np.where(df['date_died'].eq("9999-99-99"), 1, 0)
See a dummy example below,
df['date_died'] = ['9999-99-99', '9999-99-99', '2020-02-14', '2020-03-10', np.nan]
df['Survived'] = np.where(df['date_died'].eq("9999-99-99"), 1, 0)
df
date_died Survived
0 9999-99-99 1
1 9999-99-99 1
2 2020-02-14 0
3 2020-03-10 0
4 NaN 0
If you want to make Survived as 1 when date_died is np.nan, then you can change the code slightly as shown below,
df['Survived'] = np.where((df['date_died'].eq("9999-99-99")) | (df['date_died'].isna()), 1, 0)
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I have a project involving checking for pairs in a column. I´m a beginner so I tried
If a = b Then
Par = Par + 1
End If
If a = c Then
Par = Par + 1
End If
If a = d Then
Par = Par + 1
End If
etc etc but if I want this to work I´ll have to repeat this code 36 times...
Is there an easy loop for this?
Here is a simple way that might meet your needs.
Imagine a simple table in range A1:B5
A B
1 aa aa
2 cc bb
3 ee cc
4 gg dd
5 hh ee
Here we have three pairs aa & aa, cc & cc, ee & ee.
This formula will count the number of pairs:
=SUMPRODUCT(COUNTIF(A1:A5,B1:B5)) // = 3
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how much string that length is n can we create from A and B that string will not contain two A consecutively.
For example;
if n=1 , strings are A ve B => it has 2 solutions
if n=2 , strings are AB, BA, BB => it has 3 solutions
in n=3 , strings are ABB, ABA, BAB, BBA, BBB => it has 5 solutions ... etc.
I need a decrease and conquer algorithm , can you help me please ?
if n=1, result[1] = 2
if n=2, result[2] = 3
if n=3, result[3] = 5
if n=4, result[4] = 8
if n=5, result[5] = 13
so, this generates a Fibonacci sequence.
result[1] = 2
result[2] = 3
result[n] = result[n-1]+result[n-2];
Closed form of Fibonacci sequence: