Parse indented text to dictionary

user_672305 0
Viewed 85

Let's say I have the following text:

text = '''
test1:
    aaa: aaa
test2:
    bbb:bbb
test3:
    ccc:ccc
    eee:eee
test2:
    ddd:ddd
'''

I would like to parse that text and create a dict in which the keys are the lines which have zero leading spaces and the value a list of remaining lines below of each key. In case there are two same keys like 'test2' then a second list should be assigned under the same key. So my end result dict should be like the following:

{
    "test1": [
            ["    aaa: aaa"],
    ],
    "test2": [
            ["    bbb: bbb"],
            ["    ddd: ddd"],
    ],
    "test3": [
        [
            ["    ccc:ccc", "    eee:eee"],
        ]
    ],
    "test": [
            ["    ddd:ddd"],
    ]
}

Util now I wrote the following code but need some help to finish it:

parsed = {}
proj = None
prev_proj = None
lines = []
for line in text.splitlines()[1:-1]:
    leading_whitespaces = len(line) - len(line.lstrip())
    if leading_whitespaces == 0 and (re.match(r'^[a-z_]', line)):
        proj = line.split(':')[0]
        if not parsed and len(lines) == 0:
            parsed[proj] = []
            parsed[proj].append(lines)
        elif len(lines) > 0 and parsed:
            parsed[prev_proj] = []
            parsed[prev_proj].append(lines)
            lines = []
            prev_proj = proj
    else:
        lines.append(line)
python-3.x python
4 Answers

You can use defaultdict from the collections module to handle the dictionary of lists.

text = '''
test1:
    aaa: aaa
test2:
    bbb:bbb
test3:
    ccc:ccc
    eee:eee
test2:
    ddd:ddd
'''

from collections import defaultdict

out = defaultdict(list)
key = None
values = []

for line in text.splitlines():
    if not line:
        continue
    # handle a new key occurring
    if not line.startswith(' '):
        # append values to current key
        if key and values:
            out[key].append(values)
            values = []
        
        key = line.strip(':')
    # roll up values
    else:
        values.append(line)
# handle the last key and values
else:
    out[key].append(values)
user_50037560

You could try something like that:

temp_dict = dict()
latest_key = ""
for line in text.splitlines():
    if line.startswith(" ") and latest_key != "":
        for i in range(len(line)):
            if line[i] != " ":
                break
        if line.replace(" ", "") != "":     # need so no empty lines get added to values
            temp_dict[latest_key].append(line[i:])
    else:
        latest_key = line.split(":")[0]
        if latest_key not in temp_dict.keys() and latest_key != "":
            temp_dict[latest_key] = list()
print(temp_dict)
user_196467380

You can do that with a bit of text parsing. If a line starts with no leading space, update the key, otherwise append the new value to the dict.

text = '''
test1:
    aaa: aaa
test2:
    bbb:bbb
test3:
    ccc:ccc
    eee:eee
test2:
    ddd:ddd
'''

from io import StringIO

out = {}
with StringIO(text) as f:
    key = None
    for line in f:
        if line.strip() == "":  # empty lines, skip
            continue
        if line[0] != " ":  # new key
            key = line.rstrip("\n").strip(":")
            continue
        else:  # new value
            value = line.rstrip()  # remove line breaks at end
            if key not in out:
                out[key] = [value]
            else:
                out[key].append(value)
out
#{'test1': ['    aaa: aaa'],
# 'test2': ['    bbb:bbb', '    ddd:ddd'],
# 'test3': ['    ccc:ccc', '    eee:eee']}
user_56803480

You could try to use groupby():

from itertools import groupby

def new_block(line): return not line.startswith(" ")

parsed = {}
lines = filter(None, text.splitlines())
for new, block in groupby(lines, key=new_block):
    block = list(block)
    if new:
        last = parsed.setdefault(block[0].rstrip(":"), [])
    else:
        last.append(block)

Result for your example text:

{'test1': [['    aaa: aaa']],
 'test2': [['    bbb:bbb'], ['    ddd:ddd']],
 'test3': [['    ccc:ccc', '    eee:eee']]}
user_143112630
Related