CODEWITHSUNDEEP
pythondictionary
derirative23
derirative23

Reputation: 496

Splitting String to dictionary in Python

I would like to split the string to dictionary. String is taken by using 

$ sudo btmgmt find |grep rssi |sort -n |uniq -w 33

and my result is 

hci0 dev_found: 40:43:42:B3:71:11 type LE Random rssi -53 flags 0x0000 
hci0 dev_found: 44:DA:5F:EA:C6:CF type LE Random rssi -78 flags 0x0000

The goal is creating dictionary where key is MAC address and value is rssi value

dict = {
    "40:43:42:B3:71:11": "-53 ",
    "44:DA:5F:EA:C6:CF": "-78",
   }

I tried a lot of replace function to replace these Strings to empty Strings:

But it must be cleaner and better way to do this dictionary and I do not see this solution.

Any ideas?

Upvotes: 3

Views: 106

Answers (4)

Leo Skhrnkv
Leo Skhrnkv

Reputation: 1703

Input: a file with given lines 

import re
d = dict()
c1 = re.compile("((\d|\w){1,2}:){5}(\d|\w){1,2}")
c2 = re.compile(r"(-\d{2}) flags")
with open("your_saved_lines.txt") as fh:
  for l in fh.readlines():
    m1 = c1.search(l)
    m2 = c2.search(l) or "NA"
    if m1:
      d[m1.group()] = m2.group(1)
print(d)

Output: {'40:43:42:B3:71:11': '-53', '44:DA:5F:EA:C6:CF': '-78'}

Upvotes: 0

furas
furas

Reputation: 142641

If every line has the same construction then you can use split() to split text into lines and later to split every line into "words" which you can use to create element in dictionary:

s = """
hci0 dev_found: 40:43:42:B3:71:11 type LE Random rssi -53 flags 0x0000 
hci0 dev_found: 44:DA:5F:EA:C6:CF type LE Random rssi -78 flags 0x0000
"""

d = dict()

for line in s.split('\n'):
    line = line.strip() # clear spaces and enters
    if line: # skip empty lines
        words = line.split(' ')
        d[words[2]] = words[7]

print(d)

Upvotes: 2

Flostian
Flostian

Reputation: 93

Because the columns are separated by spaces, you can use the split method:

s = """hci0 dev_found: 40:43:42:B3:71:11 type LE Random rssi -53 flags 0x0000 
hci0 dev_found: 44:DA:5F:EA:C6:CF type LE Random rssi -78 flags 0x0000"""

sdic = {}

for line in s.split('\n'):
    column = line.split(' ')
    sdic[column[2]] = column[7]

print(sdic)

Upvotes: 0

Ajax1234
Ajax1234

Reputation: 71451

You can use re.findall with regex lookaheads and lookbehinds:

import re
s = """
hci0 dev_found: 40:43:42:B3:71:11 type LE Random rssi -53 flags 0x0000 
hci0 dev_found: 44:DA:5F:EA:C6:CF type LE Random rssi -78 flags 0x0000
"""
d = dict([re.findall('(?<=dev_found:\s)[A-Z\d:]+|[\-\d]+(?=\sflags)', i) for i in filter(None, s.split('\n'))])

Output:

{'40:43:42:B3:71:11': '-53', '44:DA:5F:EA:C6:CF': '-78'}

Upvotes: 0

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