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We have nested list create dictionary key as word list and the number of times they occur as values

How do I sort a list of dictionaries by a value of the dictionary?Getting key with maximum value in dictionary?Create a dictionary with list comprehension in PythonGet key by value in dictionaryPython - stringsHow to return dictionary keys as a list in Python?Codeeval Challenge not returning correct output. (Python)Why is “1000000000000000 in range(1000000000000001)” so fast in Python 3?Why is [] faster than list()?ValueError: I/O operation on closed file.

1

Actual Code is :

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts = 
for n in big_list:
 for i in big_list:
 if n == i:
 word_counts[i] = word_counts[i] + 1


print(word_counts) 

Error is :

unhashable type: 'list' on line 8

Expected result:

['one':3 , 'two': 5, 'three': 4 ,.....] like that

So please help me find correct solution

python python-2.7

share|improve this question

edited Mar 7 at 10:36

DilbertFan

12017

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

add a comment | 

1

Actual Code is :

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts = 
for n in big_list:
 for i in big_list:
 if n == i:
 word_counts[i] = word_counts[i] + 1


print(word_counts) 

Error is :

unhashable type: 'list' on line 8

Expected result:

['one':3 , 'two': 5, 'three': 4 ,.....] like that

So please help me find correct solution

python python-2.7

share|improve this question

edited Mar 7 at 10:36

DilbertFan

12017

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

add a comment | 

Actual Code is :

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts = 
for n in big_list:
 for i in big_list:
 if n == i:
 word_counts[i] = word_counts[i] + 1


print(word_counts) 

Error is :

unhashable type: 'list' on line 8

Expected result:

['one':3 , 'two': 5, 'three': 4 ,.....] like that

So please help me find correct solution

python python-2.7

share|improve this question

edited Mar 7 at 10:36

DilbertFan

12017

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

big_list = [[['one', 'two'], ['seven', 'eight']], [['nine', 'four'], ['three', 'one']], [['two', 'eight'], ['seven', 'four']], [['five', 'one'], ['four', 'two']], [['six', 'eight'], ['two', 'seven']], [['three', 'five'], ['one', 'six']], [['nine', 'eight'], ['five', 'four']], [['six', 'three'], ['four', 'seven']]]

word_counts = 
for n in big_list:
 for i in big_list:
 if n == i:
 word_counts[i] = word_counts[i] + 1


print(word_counts) 

share|improve this question

edited Mar 7 at 10:36

DilbertFan

12017

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

share|improve this question

edited Mar 7 at 10:36

DilbertFan

12017

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

edited Mar 7 at 10:36

DilbertFan

12017

edited Mar 7 at 10:36

DilbertFan

12017

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

asked Mar 7 at 10:12

Mahendra SeerviMahendra Seervi

66

1 Answer
1

active

oldest

votes

1

You can use itertools.chain in order to flatten the nested list and collections.Counter to count how many times each element occurs:

from collections import Counter
from itertools import chain

Counter(chain(*chain(*big_list)))

Output

Counter('four': 5, 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'three': 3, 
 'five': 3, 'six': 3, 'nine': 2)

---

For a solution without imports you can do something like:

d = 
for i in big_list:
 for j in i:
 for k in j:
 if not d.get(k):
 d[k] = 1
 else:
 d[k] += 1

print(d)
# 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'nine': 2, 
# 'four': 5, 'three': 3, 'five': 3, 'six': 3

share|improve this answer

edited Mar 7 at 10:44

answered Mar 7 at 10:15

yatuyatu

14.6k41542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  can we do without using module/libaray??
  
  – Mahendra Seervi
  Mar 7 at 10:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And thanks for that
  
  – Mahendra Seervi
  Mar 7 at 10:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Let me post another alternative
  
  – yatu
  Mar 7 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks @yatu done
  
  – Mahendra Seervi
  Mar 7 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Don't forget to accept @MahendraSeervi, thanks!
  
  – yatu
  Mar 7 at 10:49
  

  
  
  
  

 | 
show 3 more comments

Your Answer

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1

You can use itertools.chain in order to flatten the nested list and collections.Counter to count how many times each element occurs:

from collections import Counter
from itertools import chain

Counter(chain(*chain(*big_list)))

Output

Counter('four': 5, 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'three': 3, 
 'five': 3, 'six': 3, 'nine': 2)

---

For a solution without imports you can do something like:

d = 
for i in big_list:
 for j in i:
 for k in j:
 if not d.get(k):
 d[k] = 1
 else:
 d[k] += 1

print(d)
# 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'nine': 2, 
# 'four': 5, 'three': 3, 'five': 3, 'six': 3

share|improve this answer

edited Mar 7 at 10:44

answered Mar 7 at 10:15

yatuyatu

14.6k41542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  can we do without using module/libaray??
  
  – Mahendra Seervi
  Mar 7 at 10:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And thanks for that
  
  – Mahendra Seervi
  Mar 7 at 10:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Let me post another alternative
  
  – yatu
  Mar 7 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks @yatu done
  
  – Mahendra Seervi
  Mar 7 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Don't forget to accept @MahendraSeervi, thanks!
  
  – yatu
  Mar 7 at 10:49
  

  
  
  
  

 | 
show 3 more comments

1

You can use itertools.chain in order to flatten the nested list and collections.Counter to count how many times each element occurs:

from collections import Counter
from itertools import chain

Counter(chain(*chain(*big_list)))

Output

Counter('four': 5, 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'three': 3, 
 'five': 3, 'six': 3, 'nine': 2)

---

For a solution without imports you can do something like:

d = 
for i in big_list:
 for j in i:
 for k in j:
 if not d.get(k):
 d[k] = 1
 else:
 d[k] += 1

print(d)
# 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'nine': 2, 
# 'four': 5, 'three': 3, 'five': 3, 'six': 3

share|improve this answer

edited Mar 7 at 10:44

answered Mar 7 at 10:15

yatuyatu

14.6k41542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  can we do without using module/libaray??
  
  – Mahendra Seervi
  Mar 7 at 10:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And thanks for that
  
  – Mahendra Seervi
  Mar 7 at 10:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Let me post another alternative
  
  – yatu
  Mar 7 at 10:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks @yatu done
  
  – Mahendra Seervi
  Mar 7 at 10:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Don't forget to accept @MahendraSeervi, thanks!
  
  – yatu
  Mar 7 at 10:49
  

  
  
  
  

 | 
show 3 more comments

You can use itertools.chain in order to flatten the nested list and collections.Counter to count how many times each element occurs:

from collections import Counter
from itertools import chain

Counter(chain(*chain(*big_list)))

Output

Counter('four': 5, 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'three': 3, 
 'five': 3, 'six': 3, 'nine': 2)

---

For a solution without imports you can do something like:

d = 
for i in big_list:
 for j in i:
 for k in j:
 if not d.get(k):
 d[k] = 1
 else:
 d[k] += 1

print(d)
# 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'nine': 2, 
# 'four': 5, 'three': 3, 'five': 3, 'six': 3

share|improve this answer

edited Mar 7 at 10:44

answered Mar 7 at 10:15

yatuyatu

14.6k41542

from collections import Counter
from itertools import chain

Counter(chain(*chain(*big_list)))

Counter('four': 5, 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'three': 3, 
 'five': 3, 'six': 3, 'nine': 2)

d = 
for i in big_list:
 for j in i:
 for k in j:
 if not d.get(k):
 d[k] = 1
 else:
 d[k] += 1

print(d)
# 'one': 4, 'two': 4, 'seven': 4, 'eight': 4, 'nine': 2, 
# 'four': 5, 'three': 3, 'five': 3, 'six': 3

share|improve this answer

edited Mar 7 at 10:44

answered Mar 7 at 10:15

yatuyatu

14.6k41542

answered Mar 7 at 10:15

yatuyatu

14.6k41542

answered Mar 7 at 10:15

yatuyatu

14.6k41542