On Sun, Jun 11, 2017 at 9:54 PM, syed zaidi <syedzaid...@hotmail.co.uk> wrote:
> Thanks
> One reason fornsharing the code was that I have to manually create over 100
> variables
Don't do that.
Anytime you have to manually repeat things over and over is a sign
that you need a loop structure of some sort.
Let's look at a little code.
> with open("C:/Users/INVINCIBLE/Desktop/T2D_ALL_blastout_batch.txt", 'r') as f:
[code cut ]
> if dictionary.get(sample_name):
> dictionary[sample_name].append(operon)
> else:
> dictionary[sample_name] = []
> dictionary[sample_name].append(operon)
> locals().update(dictionary) ## converts dictionary keys to variables
Ah. That last statement there is really big warning sign. Do *not*
use locals(). locals() is almost never a good thing to use. It's
dangerous and leads to suffering.
Instead, just stick with your dictionary, and use the dictionary. If
you need a mapping from a sample name to an array of strings,
construct another dictionary to hold that information. Then most of
the code here:
> for i in main_op_list_np:
> if i in DLF002: DLF002_1.append('1')
> else:DLF002_1.append('0')
> if i in DLF004: DLF004_1.append('1')
> else:DLF004_1.append('0')
> if i in DLF005: DLF005_1.append('1')
> else:DLF005_1.append('0')
> if i in DLF006: DLF006_1.append('1')
> else:DLF006_1.append('0')
> if i in DLF007: DLF007_1.append('1')
> else:DLF007_1.append('0')
> if i in DLF008: DLF008_1.append('1')
> else:DLF008_1.append('0')
...
will dissolve into a simple dictionary lookup, followed by an array append.
Just to compare to a similar situation, consider the following. Let's
say that we want to compute the letter frequency of a sentence.
Here's one way we could do it:
########################
def histogram(message):
a = 0
b = 0
c = 0
d = 0
# .... cut
for letter in message:
if letter == 'a':
a = a + 1
else if letter == 'b':
b = b + 1
# ... cut
return {
'a': a,
'b': b,
'c': c,
# ... cut
}
########################
This is only a sketch. We can see how this would go, if we fill in
the '...' with the obvious code. But it would also be a very bad
approach. It's highly repetitive, and easy to mistpe: you might miss
a letr.
There's a much better approach. We can use a dictionary that maps
from a letter to its given frequency.
#########################
def histogram(message):
frequencies = {}
for letter in message:
frequencies[letter] = frequencies.get(letter, 0) + 1
return frequencies
#########################
Unlike the initial sketch, the version that takes advantage of
dictionaries is short and simple, and we can run it:
##############
>>> histogram('the quick brown fox')
{' ': 3, 'c': 1, 'b': 1, 'e': 1, 'f': 1, 'i': 1, 'h': 1, 'k': 1, 'o':
2, 'n': 1, 'q': 1, 'r': 1, 'u': 1, 't': 1, 'w': 1, 'x': 1}
>>> histogram('abacadabra')
{'a': 5, 'c': 1, 'b': 2, 'r': 1, 'd': 1}
##############
If you have questions, please feel free to ask.
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