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0

I have a Python dict stuffs with keys and values(list);

'car':['bmw','porsche','benz'] 'fruits':['banana','apple']

And I would like delete first value from cars: bmw and first value from fruits: banana

How can I access and delete them please? I have tried .pop(index), but it doesn't work...

python python-3.x dictionary

share|improve this question

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What did you call pop on?
  
  – wbadart
  Mar 27 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What have you tried so far?
  
  – dcg
  Mar 27 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wbadart stuffs.pop(1), but it wont access that list
  
  – Marcel Kopera
  Mar 27 at 23:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  del stuffs['car'][0]
  
  – juanpa.arrivillaga
  Mar 28 at 0:05
  

  
  
  
  

add a comment | 

0

I have a Python dict stuffs with keys and values(list);

'car':['bmw','porsche','benz'] 'fruits':['banana','apple']

And I would like delete first value from cars: bmw and first value from fruits: banana

How can I access and delete them please? I have tried .pop(index), but it doesn't work...

python python-3.x dictionary

share|improve this question

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What did you call pop on?
  
  – wbadart
  Mar 27 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What have you tried so far?
  
  – dcg
  Mar 27 at 23:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @wbadart stuffs.pop(1), but it wont access that list
  
  – Marcel Kopera
  Mar 27 at 23:55
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  del stuffs['car'][0]
  
  – juanpa.arrivillaga
  Mar 28 at 0:05
  

  
  
  
  

add a comment | 

I have a Python dict stuffs with keys and values(list);

'car':['bmw','porsche','benz'] 'fruits':['banana','apple']

And I would like delete first value from cars: bmw and first value from fruits: banana

How can I access and delete them please? I have tried .pop(index), but it doesn't work...

python python-3.x dictionary

share|improve this question

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

'car':['bmw','porsche','benz'] 'fruits':['banana','apple']

share|improve this question

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

share|improve this question

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

edited Mar 28 at 0:42

martineau

75.2k1111 gold badges103103 silver badges198198 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

asked Mar 27 at 23:51

Marcel KoperaMarcel Kopera

1531212 bronze badges

3 Answers
3

active

oldest

votes

1

An easy way of doing this is to use a for loop and iterate over each item in you're dictionary, and pop the first element:

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
for key in dictionary:
 dictionary[key].pop(0)

Or, as a list comprehension

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
[dictionary[i].pop(0) for i in dictionary]

These pieces of code reference the dictionary at each of it's keys ('car' and 'fruits') and then proceeds to use pop on the values indexed by these keys.

Edit:

Don't use a list comprehension if you don't intend to store the list. In the case where you are iterating over large values, you could run into memory errors due to storing a whole load of useless values. Such as in this case:

[print(i) for i in range(9823498)]

This will store 9823498 None values*, where as a for loop would not. but still achieve the same thing.

share|improve this answer

edited Mar 28 at 0:39

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do not use list-comprehensions for side-effects.
  
  – juanpa.arrivillaga
  Mar 28 at 0:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @juanpa.arrivillaga what do you mean side-effects?
  
  – Recessive
  Mar 28 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You are building a useless list with that list comprehension. This is not good practice at all; list comprehensions are not substitutes for for loops.
  
  – iz_
  Mar 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomothy32 Just to be clear so anyone who sees this understands why it is bad practice, can you elaborate?
  
  – Recessive
  Mar 28 at 0:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Take a look at this: stackoverflow.com/q/5753597
  
  – iz_
  Mar 28 at 0:22
  

  
  
  
  

 | 
show 2 more comments

4

You can create a new dictionary where you skip the first element using [1:]

stuffs = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
stuffs_new = k:v[1:] for k,v in stuffs.items()
# 'car': ['porsche', 'benz'], 'fruits': ['apple']

share|improve this answer

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks, it worked!
  
  – Marcel Kopera
  Mar 28 at 0:04
  

  
  
  
  

add a comment | 

0

You were almost there.

Use either:

del dict[key]

Or

dict.pop(key, value)

The second will remove but also leave the item available as a return

share|improve this answer

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This will delete an entire key, he only wants the first index of the value indexed by the key to be removed
  
  – Recessive
  Mar 27 at 23:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Recessive You are totally right, misinterpreted it. Even still, just the syntax of pop would be helpful for him I guess, considering it takes different arguments from what he had tried.
  
  – EngineersBox
  Mar 28 at 0:04
  

  
  
  
  

add a comment | 

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1

An easy way of doing this is to use a for loop and iterate over each item in you're dictionary, and pop the first element:

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
for key in dictionary:
 dictionary[key].pop(0)

Or, as a list comprehension

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
[dictionary[i].pop(0) for i in dictionary]

These pieces of code reference the dictionary at each of it's keys ('car' and 'fruits') and then proceeds to use pop on the values indexed by these keys.

Edit:

Don't use a list comprehension if you don't intend to store the list. In the case where you are iterating over large values, you could run into memory errors due to storing a whole load of useless values. Such as in this case:

[print(i) for i in range(9823498)]

This will store 9823498 None values*, where as a for loop would not. but still achieve the same thing.

share|improve this answer

edited Mar 28 at 0:39

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do not use list-comprehensions for side-effects.
  
  – juanpa.arrivillaga
  Mar 28 at 0:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @juanpa.arrivillaga what do you mean side-effects?
  
  – Recessive
  Mar 28 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You are building a useless list with that list comprehension. This is not good practice at all; list comprehensions are not substitutes for for loops.
  
  – iz_
  Mar 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomothy32 Just to be clear so anyone who sees this understands why it is bad practice, can you elaborate?
  
  – Recessive
  Mar 28 at 0:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Take a look at this: stackoverflow.com/q/5753597
  
  – iz_
  Mar 28 at 0:22
  

  
  
  
  

 | 
show 2 more comments

1

An easy way of doing this is to use a for loop and iterate over each item in you're dictionary, and pop the first element:

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
for key in dictionary:
 dictionary[key].pop(0)

Or, as a list comprehension

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
[dictionary[i].pop(0) for i in dictionary]

These pieces of code reference the dictionary at each of it's keys ('car' and 'fruits') and then proceeds to use pop on the values indexed by these keys.

Edit:

Don't use a list comprehension if you don't intend to store the list. In the case where you are iterating over large values, you could run into memory errors due to storing a whole load of useless values. Such as in this case:

[print(i) for i in range(9823498)]

This will store 9823498 None values*, where as a for loop would not. but still achieve the same thing.

share|improve this answer

edited Mar 28 at 0:39

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do not use list-comprehensions for side-effects.
  
  – juanpa.arrivillaga
  Mar 28 at 0:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @juanpa.arrivillaga what do you mean side-effects?
  
  – Recessive
  Mar 28 at 0:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You are building a useless list with that list comprehension. This is not good practice at all; list comprehensions are not substitutes for for loops.
  
  – iz_
  Mar 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Tomothy32 Just to be clear so anyone who sees this understands why it is bad practice, can you elaborate?
  
  – Recessive
  Mar 28 at 0:19
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  Take a look at this: stackoverflow.com/q/5753597
  
  – iz_
  Mar 28 at 0:22
  

  
  
  
  

 | 
show 2 more comments

An easy way of doing this is to use a for loop and iterate over each item in you're dictionary, and pop the first element:

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
for key in dictionary:
 dictionary[key].pop(0)

Or, as a list comprehension

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
[dictionary[i].pop(0) for i in dictionary]

These pieces of code reference the dictionary at each of it's keys ('car' and 'fruits') and then proceeds to use pop on the values indexed by these keys.

Edit:

Don't use a list comprehension if you don't intend to store the list. In the case where you are iterating over large values, you could run into memory errors due to storing a whole load of useless values. Such as in this case:

[print(i) for i in range(9823498)]

This will store 9823498 None values*, where as a for loop would not. but still achieve the same thing.

share|improve this answer

edited Mar 28 at 0:39

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
for key in dictionary:
 dictionary[key].pop(0)

dictionary = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
[dictionary[i].pop(0) for i in dictionary]

share|improve this answer

edited Mar 28 at 0:39

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

answered Mar 27 at 23:57

RecessiveRecessive

54544 silver badges1818 bronze badges

4

You can create a new dictionary where you skip the first element using [1:]

stuffs = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
stuffs_new = k:v[1:] for k,v in stuffs.items()
# 'car': ['porsche', 'benz'], 'fruits': ['apple']

share|improve this answer

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks, it worked!
  
  – Marcel Kopera
  Mar 28 at 0:04
  

  
  
  
  

add a comment | 

4

You can create a new dictionary where you skip the first element using [1:]

stuffs = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
stuffs_new = k:v[1:] for k,v in stuffs.items()
# 'car': ['porsche', 'benz'], 'fruits': ['apple']

share|improve this answer

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thanks, it worked!
  
  – Marcel Kopera
  Mar 28 at 0:04
  

  
  
  
  

add a comment | 

You can create a new dictionary where you skip the first element using [1:]

stuffs = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
stuffs_new = k:v[1:] for k,v in stuffs.items()
# 'car': ['porsche', 'benz'], 'fruits': ['apple']

share|improve this answer

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

stuffs = 'car':['bmw','porsche','benz'], 'fruits':['banana','apple']
stuffs_new = k:v[1:] for k,v in stuffs.items()
# 'car': ['porsche', 'benz'], 'fruits': ['apple']

share|improve this answer

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

answered Mar 27 at 23:55

SheldoreSheldore

22k55 gold badges1515 silver badges3737 bronze badges

0

You were almost there.

Use either:

del dict[key]

Or

dict.pop(key, value)

The second will remove but also leave the item available as a return

share|improve this answer

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This will delete an entire key, he only wants the first index of the value indexed by the key to be removed
  
  – Recessive
  Mar 27 at 23:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Recessive You are totally right, misinterpreted it. Even still, just the syntax of pop would be helpful for him I guess, considering it takes different arguments from what he had tried.
  
  – EngineersBox
  Mar 28 at 0:04
  

  
  
  
  

add a comment | 

0

You were almost there.

Use either:

del dict[key]

Or

dict.pop(key, value)

The second will remove but also leave the item available as a return

share|improve this answer

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  This will delete an entire key, he only wants the first index of the value indexed by the key to be removed
  
  – Recessive
  Mar 27 at 23:59
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Recessive You are totally right, misinterpreted it. Even still, just the syntax of pop would be helpful for him I guess, considering it takes different arguments from what he had tried.
  
  – EngineersBox
  Mar 28 at 0:04
  

  
  
  
  

add a comment | 

You were almost there.

Use either:

del dict[key]

Or

dict.pop(key, value)

The second will remove but also leave the item available as a return

share|improve this answer

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges

del dict[key]

dict.pop(key, value)

share|improve this answer

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges

answered Mar 27 at 23:55

EngineersBoxEngineersBox

5811 silver badge55 bronze badges