python - Efficient way to remove keys with empty strings from a dict

Question

I have a dict and would like to remove all the keys for which there are empty value strings.

metadata = {u'Composite:PreviewImage': u'(Binary data 101973 bytes)',
            u'EXIF:CFAPattern2': u''}

What is the best way to do this?

Answer 1

Python 2.X

dict((k, v) for k, v in metadata.iteritems() if v)

---

Python 2.7 - 3.X

{k: v for k, v in metadata.items() if v}

Note that all of your keys have values. It's just that some of those values are the empty string. There's no such thing as a key in a dict without a value; if it didn't have a value, it wouldn't be in the dict.

Answer 2

It can get even shorter than BrenBarn's solution (and more readable I think)

{k: v for k, v in metadata.items() if v}

Tested with Python 2.7.3.

Answer 3

If you really need to modify the original dictionary:

empty_keys = [k for k,v in metadata.iteritems() if not v]
for k in empty_keys:
    del metadata[k]

Note that we have to make a list of the empty keys because we can't modify a dictionary while iterating through it (as you may have noticed). This is less expensive (memory-wise) than creating a brand-new dictionary, though, unless there are a lot of entries with empty values.

Answer 4

If you want a full-featured, yet succinct approach to handling real-world data structures which are often nested, and can even contain cycles, I recommend looking at the remap utility from the boltons utility package.

After pip install boltons or copying iterutils.py into your project, just do:

from boltons.iterutils import remap

drop_falsey = lambda path, key, value: bool(value)
clean = remap(metadata, visit=drop_falsey)

This page has many more examples, including ones working with much larger objects from Github's API.

It's pure-Python, so it works everywhere, and is fully tested in Python 2.7 and 3.3+. Best of all, I wrote it for exactly cases like this, so if you find a case it doesn't handle, you can bug me to fix it right here.

Answer 5

BrenBarn's solution is ideal (and pythonic, I might add). Here is another (fp) solution, however:

from operator import itemgetter
dict(filter(itemgetter(1), metadata.items()))

Answer 6

Based on Ryan's solution, if you also have lists and nested dictionaries:

For Python 2:

def remove_empty_from_dict(d):
    if type(d) is dict:
        return dict((k, remove_empty_from_dict(v)) for k, v in d.iteritems() if v and remove_empty_from_dict(v))
    elif type(d) is list:
        return [remove_empty_from_dict(v) for v in d if v and remove_empty_from_dict(v)]
    else:
        return d

For Python 3:

def remove_empty_from_dict(d):
    if type(d) is dict:
        return dict((k, remove_empty_from_dict(v)) for k, v in d.items() if v and remove_empty_from_dict(v))
    elif type(d) is list:
        return [remove_empty_from_dict(v) for v in d if v and remove_empty_from_dict(v)]
    else:
        return d

Answer 7

If you have a nested dictionary, and you want this to work even for empty sub-elements, you can use a recursive variant of BrenBarn's suggestion:

def scrub_dict(d):
    if type(d) is dict:
        return dict((k, scrub_dict(v)) for k, v in d.iteritems() if v and scrub_dict(v))
    else:
        return d

Answer 8

For python 3

dict((k, v) for k, v in metadata.items() if v)

Answer 9

Quick Answer (TL;DR)

Example01

### example01 -------------------

mydict  =   { "alpha":0,
              "bravo":"0",
              "charlie":"three",
              "delta":[],
              "echo":False,
              "foxy":"False",
              "golf":"",
              "hotel":"   ",                        
            }
newdict =   dict([(vkey, vdata) for vkey, vdata in mydict.iteritems() if(vdata) ])
print newdict

### result01 -------------------
result01 ='''
{'foxy': 'False', 'charlie': 'three', 'bravo': '0'}
'''

Detailed Answer

Problem

• Context: Python 2.x
• Scenario: Developer wishes modify a dictionary to exclude blank values
  • aka remove empty values from a dictionary
  • aka delete keys with blank values
  • aka filter dictionary for non-blank values over each key-value pair

Solution

• example01 use python list-comprehension syntax with simple conditional to remove "empty" values

Pitfalls

• example01 only operates on a copy of the original dictionary (does not modify in place)
• example01 may produce unexpected results depending on what developer means by "empty"
  • Does developer mean to keep values that are falsy?
  • If the values in the dictionary are not gauranteed to be strings, developer may have unexpected data loss.
  • result01 shows that only three key-value pairs were preserved from the original set

Alternate example

• example02 helps deal with potential pitfalls
• The approach is to use a more precise definition of "empty" by changing the conditional.
• Here we only want to filter out values that evaluate to blank strings.
• Here we also use .strip() to filter out values that consist of only whitespace.

Example02

### example02 -------------------

mydict  =   { "alpha":0,
              "bravo":"0",
              "charlie":"three",
              "delta":[],
              "echo":False,
              "foxy":"False",
              "golf":"",
              "hotel":"   ",
            }
newdict =   dict([(vkey, vdata) for vkey, vdata in mydict.iteritems() if(str(vdata).strip()) ])
print newdict

### result02 -------------------
result02 ='''
{'alpha': 0,
  'bravo': '0', 
  'charlie': 'three', 
  'delta': [],
  'echo': False,
  'foxy': 'False'
  }
'''

See also

• list-comprehension
• falsy
• checking for empty string
• modifying original dictionary in place
• dictionary comprehensions
• pitfalls of checking for empty string

Answer 10

Building on the answers from patriciasz and nneonneo, and accounting for the possibility that you might want to delete keys that have only certain falsy things (e.g. '') but not others (e.g. 0), or perhaps you even want to include some truthy things (e.g. 'SPAM'), then you could make a highly specific hitlist:

unwanted = ['', u'', None, False, [], 'SPAM']

Unfortunately, this doesn't quite work, because for example 0 in unwanted evaluates to True. We need to discriminate between 0 and other falsy things, so we have to use is:

any([0 is i for i in unwanted])

...evaluates to False.

Now use it to del the unwanted things:

unwanted_keys = [k for k, v in metadata.items() if any([v is i for i in unwanted])]
for k in unwanted_keys: del metadata[k]

If you want a new dictionary, instead of modifying metadata in place:

newdict = {k: v for k, v in metadata.items() if not any([v is i for i in unwanted])}

Answer 11

I read all replies in this thread and some referred also to this thread: Remove empty dicts in nested dictionary with recursive function

I originally used solution here and it worked great:

Attempt 1: Too Hot (not performant or future-proof):

def scrub_dict(d):
    if type(d) is dict:
        return dict((k, scrub_dict(v)) for k, v in d.iteritems() if v and scrub_dict(v))
    else:
        return d

But some performance and compatibility concerns were raised in Python 2.7 world:

1. use isinstance instead of type
2. unroll the list comp into for loop for efficiency
3. use python3 safe items instead of iteritems

Attempt 2: Too Cold (Lacks Memoization):

def scrub_dict(d):
    new_dict = {}
    for k, v in d.items():
        if isinstance(v,dict):
            v = scrub_dict(v)
        if not v in (u'', None, {}):
            new_dict[k] = v
    return new_dict

DOH! This is not recursive and not at all memoizant.

Attempt 3: Just Right (so far):

def scrub_dict(d):
    new_dict = {}
    for k, v in d.items():
        if isinstance(v,dict):
            v = scrub_dict(v)
        if not v in (u'', None, {}):
            new_dict[k] = v
    return new_dict

Answer 12

Dicts mixed with Arrays

• The answer at Attempt 3: Just Right (so far) from BlissRage's answer does not properly handle arrays elements. I'm including a patch in case anyone needs it. The method is handles list with the statement block of if isinstance(v, list):, which scrubs the list using the original scrub_dict(d) implementation.

    @staticmethod
    def scrub_dict(d):
        new_dict = {}
        for k, v in d.items():
            if isinstance(v, dict):
                v = scrub_dict(v)
            if isinstance(v, list):
                v = scrub_list(v)
            if not v in (u'', None, {}):
                new_dict[k] = v
        return new_dict

    @staticmethod
    def scrub_list(d):
        scrubbed_list = []
        for i in d:
            if isinstance(i, dict):
                i = scrub_dict(i)
            scrubbed_list.append(i)
        return scrubbed_list

Answer 13

"As I also currently write a desktop application for my work with Python, I found in data-entry application when there is lots of entry and which some are not mandatory thus user can left it blank, for validation purpose, it is easy to grab all entries and then discard empty key or value of a dictionary. So my code above a show how we can easy take them out, using dictionary comprehension and keep dictionary value element which is not blank. I use Python 3.8.3

data = {'':'', '20':'', '50':'', '100':'1.1', '200':'1.2'}

dic = {key:value for key,value in data.items() if value != ''}

print(dic)

{'100': '1.1', '200': '1.2'}

Answer 14

To preserve 0 and False values but get rid of empty values you could use:

{k: v for k, v in metadata.items() if v or v == 0 or v is False}

For a nested dict with mixed types of values you could use:

def remove_empty_from_dict(d):
  if isinstance(d, dict):
    return dict((k, remove_empty_from_dict(v)) for k, v in d.items() \
            if v or v == 0 or v is False and remove_empty_from_dict(v) is not None)
  elif isinstance(d, list):
    return [remove_empty_from_dict(v) for v in d 
            if v or v == 0 or v is False and remove_empty_from_dict(v) is not None]
  else:
    if d or d == 0 or d is False:
      return d

Answer 15

An alternative way you can do this, is using dictionary comprehension. This should be compatible with 2.7+

result = {
    key: value for key, value in
    {"foo": "bar", "lorem": None}.items()
    if value
}

Answer 16

Here is an option if you are using pandas:

import pandas as pd

d = dict.fromkeys(['a', 'b', 'c', 'd'])
d['b'] = 'not null'
d['c'] = ''  # empty string

print(d)

# convert `dict` to `Series` and replace any blank strings with `None`;
# use the `.dropna()` method and
# then convert back to a `dict`
d_ = pd.Series(d).replace('', None).dropna().to_dict()

print(d_)

Answer 17

Some of Methods mentioned above ignores if there are any integers and float with values 0 & 0.0

If someone wants to avoid the above can use below code(removes empty strings and None values from nested dictionary and nested list):

def remove_empty_from_dict(d):
    if type(d) is dict:
        _temp = {}
        for k,v in d.items():
            if v == None or v == "":
                pass
            elif type(v) is int or type(v) is float:
                _temp[k] = remove_empty_from_dict(v)
            elif (v or remove_empty_from_dict(v)):
                _temp[k] = remove_empty_from_dict(v)
        return _temp
    elif type(d) is list:
        return [remove_empty_from_dict(v) for v in d if( (str(v).strip() or str(remove_empty_from_dict(v)).strip()) and (v != None or remove_empty_from_dict(v) != None))]
    else:
        return d

Answer 18

metadata ={'src':'1921','dest':'1337','email':'','movile':''}
ot = {k: v for k, v in metadata.items() if v != ''}
print(f"Final {ot}")

Answer 19

Some benchmarking:

1. List comprehension recreate dict

In [7]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
   ...: dic = {k: v for k, v in dic.items() if v is not None} 
   1000000 loops, best of 7: 375 ns per loop

2. List comprehension recreate dict using dict()

In [8]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
   ...: dic = dict((k, v) for k, v in dic.items() if v is not None)
1000000 loops, best of 7: 681 ns per loop

3. Loop and delete key if v is None

In [10]: %%timeit dic = {str(i):i for i in xrange(10)}; dic['10'] = None; dic['5'] = None
    ...: for k, v in dic.items():
    ...:   if v is None:
    ...:     del dic[k]
    ...: 
10000000 loops, best of 7: 160 ns per loop

so loop and delete is the fastest at 160ns, list comprehension is half as slow at ~375ns and with a call to dict() is half as slow again ~680ns.

Wrapping 3 into a function brings it back down again to about 275ns. Also for me PyPy was about twice as fast as neet python.