list containing integers and strings in python

Question

I am new to Python. Suppose you have python dictionary, where values are lists different elements. These values can contain only integers, only strings or both. I need to find values that contain both strings and integers.

This is my solution, which works but is not very elegant.

for key,value in dict.iteritems():
        int_count=0
        len_val=len(value)
        for v in value:
            if v.isdigit():
                int_coun+=1
        if (int_count!=0 and int_count<len_chr):
            print value

I wonder if it is conceptually possible to do something like this regex:

if [0-9].* and [a-z,A-Z].* in value:
    print value 

or in other effective and elegant way.

Thanks

EDIT

Here is an example of dictionary:

dict={ 'D00733' : ['III', 'I', 'II', 'I', 'I']
       'D00734' : ['I', 'IV', '78']
       'D00735' : ['3', '7', '18']}             

What I want is:

['I', 'IV', '78']

Answer 1

Here is a solution that you can try:

import numbers
import decimal

dct = {"key1":["5", "names", 1], "Key2":[4, 5, 3, 5]}

new_dict = {}

new_dict = {a:b for a, b in dct.items() if any(i.isalpha() for i in b) and any(isinstance(i, numbers.Number) for i in b)}

Here is a solution using regex:

import re

dct = {"key1":["5", "names", 1], "Key2":[4, 5, "hi", "56"]}

for a, b in dct.items():

   new_list = ''.join(map(str, b))

   expression = re.findall(r'[a-zA-Z]', new_list)

   expression1 = re.findall(r'[0-9]', new_list)

   if len(expression) > 0 and len(expression1) > 0:
        new_dict[a] = b

print new_dict

This algorithm builds a new dictionary with the values from the previous dictionary that meet the original criteria.