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Iterating through dictionaries

How to merge two dictionaries in a single expression?How do I efficiently iterate over each entry in a Java Map?How do I sort a list of dictionaries by a value of the dictionary?What is the best way to iterate over a dictionary?How do I sort a dictionary by value?Add new keys to a dictionary?Check if a given key already exists in a dictionaryIterating over dictionaries using 'for' loopsDelete an element from a dictionaryHow to remove a key from a Python dictionary?

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1

I have a dictionary of 3 lists, each containing 3 elements, and want to alter each element within each list separately, generating 9 new dictionaries as such:

d = 1:[a,b,c], 2:[e,f,g], 3:[h,i,j]
d1 = 1:[a+k,b,c], 2:[e,f,g], 3:[h,i,j]
d2 = 1:[a, b+k, c], 2:[e,f,g], 3:[h,i,j]
...
d9 = {1:[a,b,c], 2:[e,f,g], 3:[h,i,j+k]

Is there a way this can be done without having to hardcode each dictionary separately? Further on down the code, I will need to refer to the sets of lists separately (i.e. I will input [a,b,c+k], [d,e,f], [g,h,i] into some function to get a solution), but I am happy for the data not to be in dictionary form so long as I can still refer to their sets later!

python dictionary

share|improve this question

edited Mar 9 at 1:07

Plasmid

asked Mar 9 at 1:02

PlasmidPlasmid

174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not too clear on what you are trying to do. What are you starting with and what are you expecting to get?
  
  – busybear
  Mar 9 at 1:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Basically, I have three vectors stored in a dictionary, and I need to alter each individual coordinate in each vector separately.
  
  – Plasmid
  Mar 9 at 1:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Put d1-d9 in a list?
  
  – wjandrea
  Mar 9 at 1:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wjandrea yup! but could I somehow avoid having to type out d1-d9? (its actually 42 dictionaries in the code so I'd really rather not do things manually). Looking for a way to iterate through my initial dictionary to generate d1-d9
  
  – Plasmid
  Mar 9 at 1:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Plasmid Yes, use a list comprehension.
  
  – wjandrea
  Mar 9 at 1:11
  

  
  
  
  

 | 
show 3 more comments

1

I have a dictionary of 3 lists, each containing 3 elements, and want to alter each element within each list separately, generating 9 new dictionaries as such:

d = 1:[a,b,c], 2:[e,f,g], 3:[h,i,j]
d1 = 1:[a+k,b,c], 2:[e,f,g], 3:[h,i,j]
d2 = 1:[a, b+k, c], 2:[e,f,g], 3:[h,i,j]
...
d9 = {1:[a,b,c], 2:[e,f,g], 3:[h,i,j+k]

Is there a way this can be done without having to hardcode each dictionary separately? Further on down the code, I will need to refer to the sets of lists separately (i.e. I will input [a,b,c+k], [d,e,f], [g,h,i] into some function to get a solution), but I am happy for the data not to be in dictionary form so long as I can still refer to their sets later!

python dictionary

share|improve this question

edited Mar 9 at 1:07

Plasmid

asked Mar 9 at 1:02

PlasmidPlasmid

174

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I'm not too clear on what you are trying to do. What are you starting with and what are you expecting to get?
  
  – busybear
  Mar 9 at 1:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Basically, I have three vectors stored in a dictionary, and I need to alter each individual coordinate in each vector separately.
  
  – Plasmid
  Mar 9 at 1:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Put d1-d9 in a list?
  
  – wjandrea
  Mar 9 at 1:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wjandrea yup! but could I somehow avoid having to type out d1-d9? (its actually 42 dictionaries in the code so I'd really rather not do things manually). Looking for a way to iterate through my initial dictionary to generate d1-d9
  
  – Plasmid
  Mar 9 at 1:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Plasmid Yes, use a list comprehension.
  
  – wjandrea
  Mar 9 at 1:11
  

  
  
  
  

 | 
show 3 more comments

I have a dictionary of 3 lists, each containing 3 elements, and want to alter each element within each list separately, generating 9 new dictionaries as such:

d = 1:[a,b,c], 2:[e,f,g], 3:[h,i,j]
d1 = 1:[a+k,b,c], 2:[e,f,g], 3:[h,i,j]
d2 = 1:[a, b+k, c], 2:[e,f,g], 3:[h,i,j]
...
d9 = {1:[a,b,c], 2:[e,f,g], 3:[h,i,j+k]

Is there a way this can be done without having to hardcode each dictionary separately? Further on down the code, I will need to refer to the sets of lists separately (i.e. I will input [a,b,c+k], [d,e,f], [g,h,i] into some function to get a solution), but I am happy for the data not to be in dictionary form so long as I can still refer to their sets later!

python dictionary

share|improve this question

edited Mar 9 at 1:07

Plasmid

asked Mar 9 at 1:02

PlasmidPlasmid

174

d = 1:[a,b,c], 2:[e,f,g], 3:[h,i,j]
d1 = 1:[a+k,b,c], 2:[e,f,g], 3:[h,i,j]
d2 = 1:[a, b+k, c], 2:[e,f,g], 3:[h,i,j]
...
d9 = {1:[a,b,c], 2:[e,f,g], 3:[h,i,j+k]

share|improve this question

edited Mar 9 at 1:07

Plasmid

asked Mar 9 at 1:02

PlasmidPlasmid

174

share|improve this question

edited Mar 9 at 1:07

Plasmid

asked Mar 9 at 1:02

PlasmidPlasmid

174

asked Mar 9 at 1:02

PlasmidPlasmid

174

asked Mar 9 at 1:02

PlasmidPlasmid

174

1 Answer
1

active

oldest

votes

2

You can use a list comprehension to iterate a number from 0 to 8 and generate dicts of lists using a dict comprehension that outputs list items with possibly k added based on the quotient and remainder of 3 of the number:

[k: [a + ('k' if n // 3 + 1 == k and n % 3 == i else '') for i, a in enumerate(l)] for k, l in d.items() for n in range(9)]

so that given:

d = 1:['a','b','c'], 2:['e','f','g'], 3:['h','i','j']

this returns:

[1: ['ak', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'bk', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'ck'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['ek', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'fk', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'gk'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['hk', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'ik', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'jk']]

share|improve this answer

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743

add a comment | 

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2

You can use a list comprehension to iterate a number from 0 to 8 and generate dicts of lists using a dict comprehension that outputs list items with possibly k added based on the quotient and remainder of 3 of the number:

[k: [a + ('k' if n // 3 + 1 == k and n % 3 == i else '') for i, a in enumerate(l)] for k, l in d.items() for n in range(9)]

so that given:

d = 1:['a','b','c'], 2:['e','f','g'], 3:['h','i','j']

this returns:

[1: ['ak', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'bk', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'ck'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['ek', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'fk', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'gk'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['hk', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'ik', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'jk']]

share|improve this answer

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743

add a comment | 

2

You can use a list comprehension to iterate a number from 0 to 8 and generate dicts of lists using a dict comprehension that outputs list items with possibly k added based on the quotient and remainder of 3 of the number:

[k: [a + ('k' if n // 3 + 1 == k and n % 3 == i else '') for i, a in enumerate(l)] for k, l in d.items() for n in range(9)]

so that given:

d = 1:['a','b','c'], 2:['e','f','g'], 3:['h','i','j']

this returns:

[1: ['ak', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'bk', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'ck'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['ek', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'fk', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'gk'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['hk', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'ik', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'jk']]

share|improve this answer

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743

add a comment | 

You can use a list comprehension to iterate a number from 0 to 8 and generate dicts of lists using a dict comprehension that outputs list items with possibly k added based on the quotient and remainder of 3 of the number:

[k: [a + ('k' if n // 3 + 1 == k and n % 3 == i else '') for i, a in enumerate(l)] for k, l in d.items() for n in range(9)]

so that given:

d = 1:['a','b','c'], 2:['e','f','g'], 3:['h','i','j']

this returns:

[1: ['ak', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'bk', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'ck'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['ek', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'fk', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'gk'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['hk', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'ik', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'jk']]

share|improve this answer

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743

[k: [a + ('k' if n // 3 + 1 == k and n % 3 == i else '') for i, a in enumerate(l)] for k, l in d.items() for n in range(9)]

d = 1:['a','b','c'], 2:['e','f','g'], 3:['h','i','j']

[1: ['ak', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'bk', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'ck'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['ek', 'f', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'fk', 'g'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'gk'], 3: ['h', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['hk', 'i', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'ik', 'j'],
 1: ['a', 'b', 'c'], 2: ['e', 'f', 'g'], 3: ['h', 'i', 'jk']]

share|improve this answer

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743

answered Mar 9 at 1:14

blhsingblhsing

42.5k41743