Here are some best practices and tips for bioinformatics programming:

1. Use Version Control: Use version control systems like Git to manage your code. This helps track changes, collaborate with others, and revert to previous versions if needed.
2. Document Your Code: Write clear and concise documentation for your code, including comments and docstrings. This helps others understand your code and makes it easier to maintain.
3. Use Meaningful Variable Names: Use descriptive variable names that convey the purpose of the variable. This improves readability and makes your code easier to understand.
4. Modularize Your Code: Break your code into small, reusable modules or functions. This makes your code more modular, easier to test, and helps avoid repetition.
5. Use Libraries and Packages: Leverage existing libraries and packages for common tasks in bioinformatics. This saves time and ensures that you are using well-tested and optimized code.
6. Optimize Your Code: Write efficient code by avoiding unnecessary loops, using appropriate data structures, and optimizing algorithms. This is especially important when working with large datasets.
7. Handle Errors Gracefully: Use try-except blocks to handle errors and exceptions in your code. This prevents your program from crashing and helps identify and fix issues.
8. Test Your Code: Write unit tests to ensure that your code behaves as expected. This helps catch bugs early and ensures that your code is robust and reliable.
9. Use Virtual Environments: Use virtual environments like conda or virtualenv to manage dependencies and isolate your project’s environment. This avoids conflicts between different projects.
10. Stay Updated: Keep your software and libraries up to date to benefit from the latest features, improvements, and security patches.
11. Collaborate and Seek Feedback: Collaborate with others and seek feedback on your code. This can help improve your code quality and learn new techniques.
12. Stay Organized: Organize your code and project files in a logical manner. Use meaningful directory structures and naming conventions.

Following these best practices can help you write better code, improve productivity, and build more robust and maintainable bioinformatics applications.

Examples and Exercises

Python: Strings

Strings are immutable objects representing text. To define a string, we can write:

var = “text”

or, equivalently:

var = ‘text’

To insert special characters, we need to perform escaping with a backslash :

\

path = “data\\fasta”

or use the prefix (raw):

r

path = r”data\fasta”

Here is a reference list of escape characters. You will probably only need the most obvious

\\

\n

\t

ones, like

,

and .

To create a multi-line string, we can manually place the newline character at each line:

\n

sad_joke = “Time flies like an arrow.\nFruit flies like a banana.” print sad_joke

or we can use triple quotes:

sad_joke = “””Time flies like an arrow. Fruit flies like a banana.”””

print sad_joke

Warning

interprets special characters, while terminal echo doesn’t. Try to write:

print

print path

and (from the interpreter):

path

In the rirst case, we see one slash (the escaping slash is automatically interpreted by

), in the second case we see two slashes (the escape slash is not interpreted).

print

The same if we print .

sad_joke

String-Number conversion

We can convert a number into a string with :

str()

n = 10

s = str(n)

print n, type(n) print s, type(s)

or perform the opposite conversion:

float()

int()

n = int(“123”)

q = float(“1.23”) print n, type(n) print q, type(q)

If the string doesn’t contain the correct numeric type, Python will give an error message:

int(“3.14”)

float(“ribosome”) int(“1 2 3”)

int(“fifteen”)

# Not an int

# Not a number # Not a number # Not a number

Operations

Example. Let’s concatenate two strings:

string = “one” + ” ” + “string” length = len(string)

print “the string:”, string, “is”, length, “characters long”

Another example:

string = “Python is hell!” * 1000

print “the string is”, len(string), “characters long”

Warning

We cannot concatenate strings with other types. For example:

var = 123

print “the value of var is” + var

gives an error message. Two working alternatives:

print “the value of var is” + str(123)

or:

print “the value of var is”, var

(In the second case we miss a space between and .)

is

123

Example. The operator

substring in string

, for example:

string

checks if

appears once or more times in

substring

string = “A beautiful journey”

print “A” in string # True print “beautiful” in string # True print “BEAUTIFUL” in string # False print “ul jour” in string # True print “Gengis Khan” in string # False print ” ” in string # True

print ” ” in string # False

The result is always or .

False

True

Example. To extract a substring we can use indexes:

print alphabet[len(alphabet)–1] # “z”

print alphabet[len(alphabet)] # Error

print alphabet[10000] # Error

# “klmnop”

print alphabet[10:–10]

# “vwxy”

# “vwxyz”

print alphabet[–5:–1]

print alphabet[–5:]

# “a”

# “ab”

# “abcde” # “abcde”

print alphabet[0:1]

print alphabet[0:2] print alphabet[0:5] print alphabet[:5]

# “z”

# “y”

print alphabet[–1]

print alphabet[–2]

# “a”

# “b”

print alphabet[0]

print alphabet[1]

alphabet = “abcdefghijklmnopqrstuvwxyz”

-1

-2|

-3||

|||

…

0

|1

||2

|||

#

# # #

Warning

Extraction is inclusive with respect to the first index, but exclusive with respect to the

second. In other words corresponds to:

alphabet[i:j]

alphabet[i] + alphabet[i+1] + … + alphabet[j–1]

Note that is excluded.

alphabet[j]

Warning

Extraction return a new string, leaving the original unvaried:

alphabet = “abcdefghijklmnopqrstuvwxyz”

substring = alphabet[2:–2]

print substring print alphabet

# Is unvaried

Methods

Warning

Methods return a new string, leaving the original unvaried (as with extraction):

alphabet = “abcdefghijklmnopqrstuvwxyz”

alphabet_upper = alphabet.upper() print alphabet_upper

print alphabet # Is unvaried

Example.

lower()

upper()

and

are very simple:

text = “No Yelling”

result = text.upper() print result

result = result.lower() print result

Example. variants are also simple:

strip()

text = ” one example ”

print text.strip()

print text.lstrip() print text.rstrip()

# equivalent to text.strip(” “)

# idem # idem

print text

# text is unvaried

Note that the space between

“example”

“one”

than one character to be removed:

and

is never removed. We can specify more

“AAAA one example BBBB”.strip(“AB”)

Example. The same is valid with and :

endswith()

startswith()

text = “123456789”

print text.startswith(“1”) print text.startswith(“a”)

# True

# False

print text.endswith(“56789”) # True

print text.endswith(“5ABC9”) # False

Example.

find()

returns the position of the first occurrence of a substring, or

if the

substring never occurs:

-1

text = “123456789”

print text.find(“1”)

print text.find(“56789”)

# 0

# 4

print text.find(“Q”)

# -1

Example. returns a copy of the string where a substring is replaced with another:

replace()

text = “if roses were rotten, then” print text.replace(“ro”, “gro”)

Example. Given this unformatted string of aminoacids:

sequence = “>MAnlFKLgaENIFLGrKW ”

To increase uniformity, we want to remove the convert everything to upper case:

“>”

character, remove spaces and finally

s1 = sequence.lstrip(“>”) s2 = s2.rstrip(” “)

s3 = s2.upper() print s3

Alternatively, all in one step:

print sequence.lstrip(“>”).rstrip(” “).upper()

Why does it work? Let’s write it with brackets:

print ( ( sequence.lstrip(“>”) ).rstrip(” “) ).upper()

\ / str

\ / str

\ / str

As you can see, the result of each method is a string (as and we can invoke string methods.

s1

s2

s3

Exercises

,

e

in the example above);

1. How can I:
   1. Create a string consisting of five spaces only.
   2. Check whether a string contains at least one space.
   3. Check whether a string contains exactly five (arbitrary) characters.
   4. Create an empty string, and check whether it is really empty.
   5. Create a string that contains one hundred copies of .

“is way better”

“Python is great”

• 1. Given the strings string

1

“but cell biology is way better”

, and

.

“12345”

“biology”

“but cell”

, compose them into the

• 1. Check whether the string

begins with

(the character, not the number!)

• 1. Create a string consisting of a single character . (Check whether the output matches

\

using both the echo of the interpreter and , and possibly also with )

len()

print

• 1. Check whether the string contains one or two backslashes.

“\\”

• 1. Check whether a string (of choice) begins or ends by .

\

• 1. Check whether a string (of choice) contains at least three times at the beginning

x

and/or at the end. For instance, the following strings satisfy the desideratum:

“x. xx”

“xx. x”

“xxxx. ”

# 1 + 2 >= 3

# 2 + 1 >= 3

# 4 + 0 >= 3

while these do not:

“x. x”

“…x. ”

” ”

# 1 + 1 < 3

# 0 + 0 < 3

# 0 + 0 < 3

1. Given the string:

s = “0123456789”

which of the following extractions are correct? 1.

2.

s[10]

s[9]

s[:10]

s[1000]

s[0]

s[-1]

s[1:5]

s[-1:-5]

s[-5:-1]

s[-1000]

1. Create a two-line string that contains the two following lines of text literally, including all the special characters and the implicit newline character:

never say “never”! said the sad turtle

1. Given the strings:

string = “a 1 b 2”

digit = “DIGIT”

character = “CHARACTER”

replace all the digits in the variable with the text provided by the variable ,

digit

string

and all alphabetic characters with the content of the variable . The result should look like this:

“CHARACTER DIGIT CHARACTER DIGIT”

character

You are free to use auxiliary variables to hold any intermediate results, but do not need to.

1. Given the following multi-line sequence:

chain_a = “””SSSVPSQKTYQGSYGFRLGFLHSGTAKSVTCTYSPALNKM FCQLAKTCPVQLWVDSTPPPGTRVRAMAIYKQSQHMTEVV

RRCPHHERCSDSDGLAPPQHLIRVEGNLRVEYLDDRNTFR HSVVVPYEPPEVGSDCTTIHYNYMCNSSCMGGMNRRPILT IITLEDSSGNLLGRNSFEVRVCACPGRDRRTEEENLRKKG EPHHELPPGSTKRALPNNT”””

which represents the aminoacid sequence of the DNA-binding domain of the Tumor Suppressor Protein TP53 , answer the following questions.

• 1. How many lines does it hold?
  2. How long is the sequence? (Do not forget to ignore the special characters!)
  3. Remove all newline characters, and put the result in a new variable .

sequence

• 1. How many cysteines are there in the sequence? How many histidines ?

“H”

“C”

• 1. Does the chain contain the sub-sequence ? In what position?

[i:j]

“NLRVEYLDDRN”

• 1. How can I use line from

chain_a

and the sub-string extraction

?

find()

operators to extract the first

1. Given (a small portion of) the tertiary structure of chain A of the TP53 protein:

structure_chain_a = “””SER A 96 77.253 20.522 75.007

VAL A 97 76.066 22.304 71.921

PRO A 98 77.731 23.371 68.681

SER A 99 80.136 26.246 68.973

GLN A 100 79.039 29.534 67.364

LYS A 101 81.787 32.022 68.157″””

Each line represents an \(C_\alpha\) atom of the backbone of the structure. Of each atom,

we know: – the aminoacid code of the residue – the chain (which is always in this

“A”

example) – the position of the residue within the chain (starting from the N-terminal) – and the \(x, y, z\) coordinates of the atom

• 1. Extract the second line using and the extraction operator. Put the line in a new

find()

variable .

line

• 1. Extract the coordinates of the second residue, and put them into three variables , ,

x

y

and .

z

• 1. Extract the coordinates from third residue as well, putting them in different variables

, ,

z_prime

y_prime

x_prime

• 1. Compute the Euclidean distance between the two residues:

\(d((x,y,z),(x’,y’,z’)) = \sqrt{(x-x’)^2 + (y-y’)^2 + (z-z’)^2}\)

Hint: make sure to use numbers when computing the distance.

float

1. Given the following DNA sequence, part of the BRCA2 human gene:

dna_seq = “””GGGCTTGTGGCGCGAGCTTCTGAAACTAGGCGGCAGAGGCGGAGCCGCT GTGGCACTGCTGCGCCTCTGCTGCGCCTCGGGTGTCTTTT

GCGGCGGTGGGTCGCCGCCGGGAGAAGCGTGAGGGGACAG ATTTGTGACCGGCGCGGTTTTTGTCAGCTTACTCCGGCCA AAAAAGAACTGCACCTCTGGAGCGG”””

• 1. Calculate the GC-content of the sequence
  2. Convert the DNA sequence into an RNA sequence
  3. Assuming that this sequence contains an intron ranging from nucleotide 51 to nucleotide 156, store the sequence of the intron in a string, and the sequence of the spliced transcript in another string.