Find Value in Dictionary Using Regex in Python

Python dictionary search values for keys using regular expression

If you only need to check keys that are starting with "seller_account", you don't need regex, just use startswith()

my_dict={'account_0':123445,'seller_account':454545,'seller_account_0':454676, 'seller_account_number':3433343}

for key, value in my_dict.iteritems():   # iter on both keys and values
        if key.startswith('seller_account'):
                print key, value

or in a one_liner way :

result = [(key, value) for key, value in my_dict.iteritems() if key.startswith("seller_account")]

NB: for a python 3.X use, replace iteritems() by items() and don't forget to add () for print.

how to get dict value by regex in python

>>> [x for d in dict1 for x in dict1[d] if d.startswith("s")]
[1, 2, 3, 4, 5, 6, 10, 11]

or, if it needs to be a regex

>>> regex = re.compile("^s")
>>> [x for d in dict1 for x in dict1[d] if regex.search(d)]
[1, 2, 3, 4, 5, 6, 10, 11]

What you're seeing here is a nested list comprehension. It's equivalent to

result = []
for d in dict1:
    for x in dict1[d]:
        if regex.search(d):
            result.append(x)

As such, it's a little inefficient because the regex is tested way too often (and the elements are appended one by one). So another solution would be

result = []
for d in dict1:
    if regex.search(d):
       result.extend(dict1[d])

Finding key in python dictionary using regex

If you must use regular expressions, you can use re.match while iterating over the dictionary.

import re

dictionary = {'Storgatan': 46, 'Talgvägen': 51, 'Malmstigen': 8, 'Huvudgatan': 3}

regex = '.*gatan$'

results = [v for k, v in dictionary.items() if re.match(regex, k)]

print(results)

Output:

[46, 3]

N.B: This'll be slow for large dictionaries

If you want just the key names:

matching_keys = [k for k in dictionary if re.match(regex, k)]

Python - Find key in dictionary using regex

This regex solution should work for you.

import re

# fixed pre-given dictionary
mydict ={
    'entry  1':'first',
    'entry  2':'second',
    'entry    4' : 'fourth',
    'entry 10':'tenth'}

# the number that you have been given; in your example, 4
num = 4
#loop through the keys and values of the dictionary
for key, val in mydict.items():
    # use regex to find all integers in the key. Ignores the whitespace
    ints = list(map(int, re.findall(r'\d+', key)))
    # if integer in the key, is the number that you have been given
    if num in ints:
        # then you have found your key and value
        print(f'key: {key}\nval: {val}')

It loops through every key, value pair in the dictionary and finds the key that ends with the desired number.

Insert values in dictionary using regex which includes key in the pattern

You could use

import re

data = """
Name  : Luke Cameron 
Age and Sex : 37/Male
Haemoglobin       13.0            g/dL"""

rx = re.compile(r'^(?P<key>[^:\n]+):(?P<value>.+)', re.M)

result = {}
for match in rx.finditer(data):
    key = match.group('key').rstrip()
    value = match.group('value').strip()
    try:
        key1, key2 = key.split(" and ")
        value1, value2 = value.split("/")
        result.update({key1: value1, key2: value2})
    except ValueError:
        result.update({key: value})

print(result)

Which yields

{'Name': 'Luke Cameron', 'Age': '37', 'Sex': 'Male'}

Iterating Through Dictionary To Get Value Using Regex

I think this might help

banks = [{'bank_code': 'BANKA',
  'profit_rate': 2.7},
{'bank_code': 'BANKB',
  'interest_rate': 2.6}]

for bank in banks:
    x = list(bank.keys())[1]
    print(bank[x])

Python: How to sort dictionary value using regex

The following code is one way of producing the results you're asking for using a dictionary. The only catch is that since a dictionary doesn't allow duplicate keys, you need the dictionary values to be lists (containing the duplicate key values).

Please note I had to change tab character to ||| since repl.it isn't allowing me to use the tab character in files. You would have to replace ||| with \t in the script below.

See it run here

import glob
from collections import defaultdict, OrderedDict

d = defaultdict(list)

def read():
  for files in glob.glob('*.txt'):
    with open(files, 'r') as f:
      for line in f:
        r = line.rstrip('\n').split('|||')
        d[r[2]].append(r)

def display(d):
  od = OrderedDict(sorted(d.items(), reverse=True))
  for k,v in od.items(): print(k,v)

read()
display(d)

The output for the script above is below:

900 [['file', 'GameOfThrones', '900', '0']]
504 [['file', 'DC/Batman', '504', '1']]
444 [['file', 'Science/Chemistry', '444', '1']]
342 [['file', 'Marvel/CaptainAmerica', '342', '0'], ['file', 'Math/Calculus', '342', '0']]
324 [['file', 'Psychology', '324', '1']]
300 [['file', 'Marvel/GuardiansOfGalaxy', '300', '1']]
234 [['file', 'Anthropology', '234', '0']]
200 [['file', 'DC/Superman', '200', '1'], ['file', 'Science/Biology', '200', '1']]

Different Outputs

Changing print(k,v) to another format gets you different output as you've noted in the comments below. For example:

print(k, [x[1] for x in v])

This results in the following:

900 ['GameOfThrones']
504 ['DC/Batman']
444 ['Science/Chemistry']
342 ['Marvel/CaptainAmerica', 'Math/Calculus']
324 ['Psychology']
300 ['Marvel/GuardiansOfGalaxy']
234 ['Anthropology']
200 ['DC/Superman', 'Science/Biology']

Python : find and replace patterns in the value of dictionary that is a list of strings

replace() is an inbuilt function in Python programming language that returns a copy of the string where all occurrences of a substring is replaced with another substring.

Ex.

dictionarylst = {0:["example inside some sentence", "something else", "some 
                  blah"], 1:["testing", "some other word"],2:["a new expression",
                 "my cat is cute"]}

wordslist = ["expression 1", "my expression", "other", "blah"]
dictionarycleaned = {}

def match_pattern(wordslist,value):
    new_list = []
    for text in value:
        # temp variable hold latest updated text
        temp = text
        for word in wordslist:
            if word in text:
                # replace text string with whitespace if word in text
                temp = temp.replace(word,"")
        new_list.append(temp)
    return new_list


for k,v in dictionarylst.items():
    dictionarycleaned[k] = match_pattern(wordslist, v)

print(dictionarycleaned)

O/P:

{0: ['example inside some sentence', 'something else', 'some '], 1: ['testing', 'some  
 word'], 2: ['a new expression', 'my cat is cute']}