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Answer 1

+1 A:

import re

results = []
temp = {}
for item in data:
    (key, value) = re.search('(.*?):(.*)', item).groups()
    if temp.has_key(key): temp = {}
    temp[key] = value
    if temp not in results: results.append(temp)

mkotechno 2009-12-23 04:31:11

That produces a list of lists, I would ultimately like a list of dicts for easier parsing later on.

CarpeNoctem 2009-12-23 04:34:32

edited with dicts instead lists

mkotechno 2009-12-23 04:45:58

Answer 2

+1 A:

If you have no more info than "each repetition of a key signals the need to start a new dict", your code can be improved only marginally, for example as:

results = []
curd = {}
for x in data:
  k, v = x.split(':', 1)
  if k in curd:
    results.append(curd)
    curd = {}
  curd[k] = v
results.append(curd)

i.e., no need to keep an intermediate list tmp rather than an intermediate dict curd. The semantics are subtly different -- you're initiating a new dict only when both key and value coincide (so an item such as 'Status:Borked' would "trample over" one being built from 'Status:Ok', for example), I'm taking the key only as the identifier (so, no trampling over in such a case) -- you sure the exact semantics you implement are what you require?

Alex Martelli 2009-12-23 04:42:06

Sure is a strong word that I am far from - this is my first python script. I would like a list of dicts to make iteration easier later on. I didn't want to paste in the entire script as it might have scared you helpful people away.

CarpeNoctem 2009-12-23 04:57:24

Yes, "each repetition of a key signals the need to start a new dict", since the amount of key:value elements in the list can change. This is what I was trying to write but wasn't quite there. Thanks.

CarpeNoctem 2009-12-23 05:08:07

Answer 3

+6 A:

result = [{}]
for item in data:
    key, val = item.split(":", 1)
    if key in result[-1]:
        result.append({})
    result[-1][key] = val

teepark 2009-12-23 04:43:07

Note that `result[-1]` means "the last element" - using negative indices counts from the back

Smashery 2009-12-23 04:50:06

So would this be what people speak of when they say "pythonic"? It's concise and it works perfectly.

CarpeNoctem 2009-12-23 04:56:46

Yeah, I'd describe this as "pythonic". Rather than storing temporary variables, everything is put straight into the result list.

Smashery 2009-12-23 04:59:56

Good (and readable) code, although I personally will add a couple of comments. Just a question of preference.

Khelben 2009-12-23 07:09:07

Answer 4

A:

ret = []
ITEMS_AMOUNT = 5 
while True:
        tmp = {}
        for i in data[0:ITEMS_AMOUNT]:
                tmp.update(dict([i.split(':', 1)]))
        ret.append(tmp)

        if len(data) == ITEMS_AMOUNT:
                break
        data = data[ITEMS_AMOUNT:]

print ret

nandu 2009-12-23 04:52:56

Answer 5

A:

d=dict([])
c=0
whatiwant=["ID","Status","Name","State","Failure Predicted"]
for line in open("file"):
    line=line.rstrip()
    sline=line.split(":",1)
    sline[0]=sline[0].strip()
    if sline[0]=="ID":
        c+=1
        d.setdefault(c,[])
    if sline[0] in whatiwant:
        d[c].append((sline[0],' '.join(sline[1:])))
for i,j in d.iteritems():
    print i,j

output

$ ./python.py
1 [('ID', ' 0:0:0'), ('Status', ' Ok'), ('Name', ' Physical Disk 0:0:0'), ('State', ' Online'), ('Failure Predicted', ' No')]
2 [('ID', ' 0:0:1'), ('Status', ' Ok'), ('Name', ' Physical Disk 0:0:1'), ('State', ' Online'), ('Failure Predicted', ' No')]

ghostdog74 2009-12-23 05:00:54

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Python: Split list into list of dicts?

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