I am trying to store a string variable containg some names, I want to store the respective variable in a list and print it, but am unable print the values which are stored in variable.
name='vsb','siva','anand','soubhik' #variable containg some names
lis=['name'] # storing the variable in a list
for x in lis:
print(x) #printing the list using loops
Image:
Maybe dictionary? Try this
variable_1 = "aa"
variable_2 = "bb"
lis = {}
lis['name1'] = variable_1
lis['name2'] = variable_2
for i in lis:
print(i)
print(lis[i])
Your name variable is actually a tuple.
Example of tuple declaration:
tup1 = ('physics', 'chemistry', 1997, 2000)
tup2 = (1, 2, 3, 4, 5 )
tup3 = "a", "b", "c", "d"
Example of list declaration:
list1 = ['physics', 'chemistry', 1997, 2000]
list2 = [1, 2, 3, 4, 5 ]
list3 = ["a", "b", "c", "d"]
For a better understanding you should read The Python Standard Library or do a tutorial.
For your problem maybe the dictionary is the solution:
# A tuple is a sequence of immutable Python objects
name='vsb','siva','anand','soubhik'
print('Tuple: ' + str(name)) # ('vsb', 'siva', 'anand', 'soubhik')
# This is a list containing one element: 'name'
lis=['name']
print('List: ' + str(lis)) # ['name']
# Dictionry with key 'name' and vlue ('vsb','siva','anand','soubhik')
dictionary={'name':name}
print('Dictionary: ' + str(dictionary))
print('Dictionary elements:')
print(dictionary['name'])
print('Tuple elements:')
for x in name:
print(x)
print('List elements:')
for x in lis:
print(x)
Output
Tuple: ('vsb', 'siva', 'anand', 'soubhik')
List: ['name']
Dictionary: {'name': ('vsb', 'siva', 'anand', 'soubhik')}
Dictionary elements:
('vsb', 'siva', 'anand', 'soubhik')
Tuple elements:
vsb
siva
anand
soubhik
List elements:
name
Related
I have a 2d list with arbitrary strings like this:
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
I want to create a dictionary out of this:
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
How do I do this? This answer answers for 1D list for non-repeated values, but, I have a 2d list and values can repeat. Is there a generic way of doing this?
Maybe you could use two for-loops:
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
d = {}
overall_idx = 0
for sub_lst in lst:
for word in sub_lst:
if word not in d:
d[word] = overall_idx
# Increment overall_idx below if you want to only increment if word is not previously seen
# overall_idx += 1
overall_idx += 1
print(d)
Output:
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
You could first convert the list of lists to a list using a 'double' list comprehension.
Next, get rid of all the duplicates using a dictionary comprehension, we could use set for that but would lose the order.
Finally use another dictionary comprehension to get the desired result.
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
# flatten list of lists to a list
flat_list = [item for sublist in lst for item in sublist]
# remove duplicates
ordered_set = {x:0 for x in flat_list}.keys()
# create required output
the_dictionary = {v:i for i, v in enumerate(ordered_set)}
print(the_dictionary)
""" OUTPUT
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
"""
also, with collections and itertools:
import itertools
from collections import OrderedDict
lstdict={}
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
lstkeys = list(OrderedDict(zip(itertools.chain(*lst), itertools.repeat(None))))
lstdict = {lstkeys[i]: i for i in range(0, len(lstkeys))}
lstdict
output:
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
I am trying to convert list elements of a given list to list value of another list. Here is an example:
list1 = ['apple', 'tomato', 'milk', 'beans', 'mango']
list2 = [{0,4}, {1,3}, 2]
result = []
expected output:
[{'apple', 'banana'}, {'tomato', 'beans'}, 'milk']
I am trying to use the below code, but it does not work:
for i in list2:
result.append(list1[i])
Assuming list2 will only contain set, list, or int. This would be my approach.
list1 = ['apple', 'tomato', 'milk', 'beans', 'mango']
list2 = [{0,4}, {1,3}, 2]
result = []
# [{'apple', 'banana'}, {'tomato', 'beans'}, 'milk']
for x in list2:
temp_lst = []
if type(x) in [set, list]:
new_set = set()
for y in x:
new_set.add(list1[y])
temp_lst.append(new_set)
else:
temp_lst.append(list1[x])
result.extend(temp_lst)
print(result)
I am trying to write a function to extract only words unique to each key and list them in a dictionary output like {"key1": "unique words", "key2": "unique words", ... }. I start out with a dictionary. To test with I created a simple dictionary:
d = {1:["one", "two", "three"], 2:["two", "four",
"five"], 3:["one","four", "six"]}
My output should be:
{1:"three",
2:"five",
3:"six"}
I am thinking maybe split in to separate lists
def return_unique(dct):
Klist = list(dct.keys())
Vlist = list(dct.values())
aList = []
for i in range(len(Vlist)):
for j in Vlist[i]:
if
What I'm stuck on is how do I tell Python to do this: if Vlist[i][j] is not in the rest of Vlist then aList.append(Vlist[i][j]).
Thank you.
You can try something like this:
def return_unique(data):
all_values = []
for i in data.values(): # Get all values
all_values = all_values + i
unique_values = set([x for x in all_values if all_values.count(x) == 1]) # Values which are not duplicated
for key, value in data.items(): # For Python 3.x ( For Python 2.x -> data.iteritems())
for item in value: # Comparing values of two lists
for item1 in unique_values:
if item == item1:
data[key] = item
return data
d = {1:["one", "two", "three"], 2:["two", "four", "five"], 3:["one","four", "six"]}
print (return_unique(d))
result >> {1: 'three', 2: 'five', 3: 'six'}
Since a key may have more than one unique word associated with it, it makes sense for the values in the new dictionary to be a container type object to hold the unique words.
The set difference operator returns the difference between 2 sets:
>>> a = set([1, 2, 3])
>>> b = set([2, 4, 6])
>>> a - b
{1, 3}
We can use this to get the values unique to each key. Packaging these into a simple function yields:
def unique_words_dict(data):
res = {}
values = []
for k in data:
for g in data:
if g != k:
values += data[g]
res[k] = set(data[k]) - set(values)
values = []
return res
>>> d = {1:["one", "two", "three"],
2:["two", "four", "five"],
3:["one","four", "six"]}
>>> unique_words_dict(d)
{1: {'three'}, 2: {'five'}, 3: {'six'}}
If you only had to do this once, then you might be interested in the less efficeint but more consice dictionary comprehension:
>>> from functools import reduce
>>> {k: set(d[k]) - set(reduce(lambda a, b: a+b, [d[g] for g in d if g!=k], [])) for k in d}
{1: {'three'}, 2: {'five'}, 3: {'six'}}
I have two lists with some items in common and some not. I would like to compare the two lists and get the count of items that matched.
list1 = ['apple','orange','mango','cherry','banana','kiwi','tomato','avocado']
list2 = ['orange','avocado','kiwi','mango','grape','lemon','tomato']
Pls advice how to do this in python
Use Counters and dictionary comprehension.
list1 = ['apple','orange','mango','cherry','banana','kiwi','tomato','avocado']
list2 = ['orange','avocado','kiwi','mango','grape','lemon','tomato']
c1 = Counter(list1)
c2 = Counter(list2)
matching = {k: c1[k]+c2[k] for k in c1.keys() if k in c2}
print(matching)
print('{} items were in both lists'.format(len(macthing))
Output:
{'avocado': 2, 'orange': 2, 'tomato': 2, 'mango': 2, 'kiwi': 2}
5 items were in both lists
I think you can use set.intersection within a comprehension like this example:
list1 = ['apple','orange','mango','cherry','banana','kiwi','tomato','avocado']
list2 = ['orange','avocado','kiwi','mango','grape','lemon','tomato']
result = {elm: list1.count(elm) + list2.count(elm) for elm in set.intersection(set(list1), set(list2))}
Output:
{'kiwi': 2, 'avocado': 2, 'orange': 2, 'tomato': 2, 'mango': 2}
How can I print a nested python dictionary in a specific format?
So, my dictionary is looks like this:
dictionary = {'Doc1':{word1: 3, word2: 1}, 'Doc2':{word1: 1, word2: 14, word3: 3}, 'Doc3':{word1: 2}}
I tried the following way:
for x, y in dictionary.items():
print(x,":", y)
But it will printL`
Doc1: {word1:3, word2: 1}
Doc2: {word1:1, word2:14, word3:3}
Doc3: {word1:2}
How to get rid of the bracket and print the plain information?
I want to print on the following format:
Doc1: word1:3; word2:1
Doc2: word1:1; word2:14; word3: 3
Doc3: word1:2;
:
in your case 'y' is a dict, so if you want to print it differently you can override the repr (representation of the object) or dict.
alternatively you can use some recursion here
def print_d(dd):
if type(dd) != dict:
return str(dd)
else:
res = []
for x,y in dd.items():
res.append(''.join((str(x),':',print_d(y))))
return '; '.join(res)
if __name__=='__main__':
dictionary = {'Doc1':{'word1': 3, 'word2': 1}, 'Doc2':{'word1': 1, 'word2': 14, 'word3': 3}, 'Doc3':{'word1': 2}}
for x, y in dictionary.items():
print(x,": ", print_d(y))
Aside from the fact that your original dictionary declaration is not valid python unless each word is a defined variable, this seems to work:
import json
print(json.dumps(dictionary).replace("{","").replace(',','').replace("}","\n").replace('"',''))
Result:
Doc1: word1: 3 word2: 1
Doc2: word1: 1 word2: 14 word3: 3
Doc3: word1: 2