CODEWITHSUNDEEP
python
oaklander114
oaklander114

Reputation: 3383

split values in dictionary in separate values

I have this type of string:

sheet = """
magenta
turquoise,PF00575
tan,PF00154,PF06745,PF08423,PF13481,PF14520
turquoise, PF00011
NULL
"""

Every line starts with an identifier (e.g. tan, magenta...) What I want is to count the number of occurrences of each PF-number per identifier.

So, the final structure would be something like this:

         magenta  turquoise tan NULL
PF00575   0          0       0   0
PF00154   0          1       0   0
PF06745   0          0       1   0
PF08423   0          0       1   0
PF13481   0          0       1   0
PF14520   0          0       1   0
PF00011   0          1       0   0

I started with making a a dictionary where every first word on a line is a key and then I want as values the PF-numbers behind it.

When I use this code, I get the values as a list of strings instead of as separate values in the dictionary:

lines = []
lines.append(sheet.split("\n"))
flattened=[]
flattened = [val for sublist in lines for val in sublist]
pfams = []
for i in flattened:
    pfams.append(i.split(","))
d = defaultdict(list)
for i in pfams:
pfam = i[0]
d[pfam].append(i[1:])

So, the result is this:

defaultdict(<type 'list'>, {'': [[], []], 'magenta': [[]], 'NULL': [[]], 'turquoise': [['PF00575']], 'tan': [['PF00154', 'PF06745', 'PF08423', 'PF13481', 'PF14520']]})

How can I split up the PFnumbers so that they are separate values in the dictionary and then count the number of occurrences of each unique PF-number per key?

Upvotes: 0

Views: 100

Answers (2)

oaklander114
oaklander114

Reputation: 3383

With thanks to dwblas on devshed, this is the most efficient way I've found to tackle the task:

I build a dictionary whose key is the PFnumber, and a list ordered by how I want the colors printed.

colors_list= ['cyan','darkorange','greenyellow','yellow','magenta','blue','green','midnightblue','brown','darkred','lightcyan','lightgreen','darkgreen','royalblue','orange','purple','tan','grey60','darkturquoise','red','lightyellow','darkgrey','turquoise','salmon','black','pink','grey','null']
lines = sheet.splitlines()
counts = {}

for line in lines:
    parts = line.split(",")
    if len(parts) > 1:
        ## doesn't break out the same item in the list many times
        color=parts[0].strip().lower()
        for key in parts[1:]:  ## skip color
            key=key.strip()
            if key not in counts:
                ## new key and list of zeroes-print it if you want to verify
                counts[key]=[0 for ctr in range(len(colors_list))]

            ## offset number/location of this color in list
            el_number=colors_list.index(color)
            if color > -1:  ## color found
                counts[key][el_number] += 1
            else:
                print "some error message"

import csv

with open("out.csv", "wb") as f:
    writer=csv.writer(f)
    writer.writerow( ["PFAM",] + colors_list)
    for pfam in counts:
    writer.writerow([pfam] + counts[pfam])

Upvotes: 0

emvee
emvee

Reputation: 4449

Use collections.Counter (https://docs.python.org/2/library/collections.html#collections.Counter)

import collections

sheet = """
magenta
turquoise,PF00575
tan,PF00154,PF06745,PF08423,PF13481,PF14520
NULL
"""

acc = {}
for line in sheet.split('\n'):
    if line == "NULL":
         continue
    parts = line.split(',')
    acc[parts[0]] = collections.Counter(parts[1])

EDIT: Now with accumulating all PF values for each key

acc = collections.defaultdict(list)
for line in sheet.split('\n'):
    if line == "NULL":
         continue
    parts = line.split(',')
    acc[parts[0]] += parts[1:]
acc = {k: collections.Counter(v) for k,v in acc.iteritems()}

Final edit Count the occurrence of colours per PF value, which is what we were after all along, in the end:

acc = collections.defaultdict(list)
for line in sheet.split('\n'):
    if line == "NULL":
         continue
    parts = line.split(',')
    for pfval in parts[1:]
         acc[ pfval ] += [ parts[0] ]
acc = {k: collections.Counter(v) for k,v in acc.iteritems()}

Upvotes: 1

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