Update the value of a dictionary with elements in the lists Python

Question

I've created a list and a dictionary for calculating amino acid frequencies.

I want to basically replace the values in the dictionary with values in the list.

So for example,

what I have now is.

L1 = [0.0429, 0.0071, 0.05, 0.0929, 0.0429, 0.0143, 0.0071, 0.0429, 0.0929, 0.1143, 0.0643, 0.0643, 0.05, 0.0286, 0.0286, 0.0643, 0.0857, 0.0714, 0.0143, 0.0214]

and

D1 = OrderedDict([('A', 6), ('C', 1), ('D', 7), ('E', 13), ('F', 6), ('G', 2), ('H', 1), ('I', 6), ('K', 13), ('L', 16), ('M', 9), ('N', 9), ('P', 7), ('Q', 4), ('R', 4), ('S', 9), ('T', 12), ('V', 10), ('W', 2), ('Y', 3)])

I want to replace every value in D1 to values in L1. So for example I want ([('A', 0.0429),('C',0.0071).... and so on.

This is the code that creates the list (L1) and the dictionary(D1).

for seq_record in SeqIO.parse(f1,'fasta'):

        sorted_dict = (collections.OrderedDict(sorted(Counter(seq_record.seq).items())))
        total = float(sum(Counter(seq_record.seq).values()))
        print sorted_dict
        aa_frequency =(round(value/total,4) for key, value in sorted_dict.items())
        aa_frequency_value = []
        for i in aa_frequency:
            aa_frequency_value.append(i)

Adam Smith · Accepted Answer · 2015-11-25 22:35:17Z

3

Pretty straightforward since you already have the values and keys in order.

for val, key in zip(L1, D1):
    D1[key] = val

Seems like you should probably just do this in-place, since you're not adding or deleting elements

total = float(um(Counter(seq_record.seq).values()))
for key,val in sorted_dict.items():
    newval = round(val/total, 4)
    sorted_dict[key] = newval

answered Nov 25, 2015 at 22:35

Adam Smith

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1 Comment

Danny Over a year ago

Thanks this is exactly what I wanted,

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