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Get averages from a python dictionary for example if i have the next dictionary:
students={'Dan':(5,8,8), 'Tim':(7), 'Richard':(9,9)}
And i would like to print de dictionary in the next form:
results={'Dan':(7), 'Tim':(7), 'Richard':(9)}
is there any function that i can use? Im new coding in python so dictionaries are a bit confusing for me.
If you want that the avg values will be a tuple element (I don't see any reason to do so but maybe I don't have enough context), try:
results={k: (sum(v)/len(v),) for k,v in students.items()}
I'd do:
result = {}
for k, v in students.items():
if type(v) in [float, int]:
result[k] = v
else:
result[k] = sum(v) / len(v)
I was trying this but realized we had a problem summing a tuple of length 1. So you can do it this way.
results = {}
for k, v in students.items():
print(v)
if (isinstance(v, int)):
results[k] = v
else:
results[k] = sum(v) / len(v)
A pythonic solution would be to use dictionary comprehension to create the results dictionary.
def avg(l):
return sum([l]) / len(l)
results = {key: (avg(val)) for (key, val) in students.items()}
You need brackets around the l in sum so the tuple is treated as a list. I would further change the data structure to a list instead of tuple.
I have two dictionaries created this way :
tr = defaultdict(list)
tr = { 'critic' : '2_critic',
'major' : '3_major',
'all' : ['2_critic','3_major']
}
And the second one :
scnd_dict = defaultdict(list)
And contains values like this :
scnd_dict = {'severity': ['all']}
I want to have a third dict that will contain the key of scnd_dict and its corresponding value from tr.
This way, I will have :
third_dict = {'severity' : ['2_critic','3_major']}
I tried this, but it didn't work :
for (k,v) in scnd_dict.iteritems() :
if v in tr:
third_dict[k].append(tr[v])
Any help would be appreciated. Thanks.
Well...
from collections import defaultdict
tr = {'critic' : '2_critic',
'major' : '3_major',
'all' : ['2_critic','3_major']}
scnd_dict = {'severity': ['all']}
third_dict = {}
for k, v in scnd_dict.iteritems():
vals = []
if isinstance(v, list):
for i in v:
vals.append(tr.get(i))
else:
vals.append(tr.get(v))
if not vals:
continue
third_dict[k] = vals
print third_dict
Results:
>>>
{'severity': [['2_critic', '3_major']]}
Will do what you want. But I question the logic of using defaultdicts here, or of have your index part of a list...
If you use non-lists for scnd_dict then you can do the whole thing much easier. Assuming scnd_dict looks like this: scnd_dict = {'severity': 'all'}:
d = dict((k, tr.get(v)) for k, v in scnd_dict.items())
# {'severity': ['2_critic', '3_major']}
Your problem is that v is a list, not an item of a list. So, the if v in tr: will be false. Change your code so that you iterate over the items in v
third_dict = {k: [t for m in ks for t in tr[m]] for k,ks in scnd_dict.iteritems()}
The second dict's value is list, not str, so the code blow will work
for (k, v) in send_dict.iteritems():
if v[0] in tr.keys():
third_dict[k] = tr[v[0]]
The problem is that the third dictionary does not knows that the values is a list
for k in scnd_dict:
for v in scnd_dict[k]:
print v
for k2 in tr:
if v==k2:
if k not in third_dict:
third_dict[k]=tr[k2]
else:
third_dict[k]+=tr[k2]
third_dict = {k: tr[v[0]] for k, v in scnd_dict.iteritems() if v[0] in tr}
This
tr = defaultdict(list)
is a waste of time if you are just rebinding tr on the next line. Likewise for scnd_dict.
It's a better idea to make all the values of tr lists - even if they only have one item. It will mean less special cases to worry about later on.
Following works for a dictionary, but not OrderedDict. For od it seems to form an infinite loop. Can you tell me why?
If the function input is dict it has to return dict, if input is OrderedDict it has to return od.
def key_lower(d):
"""returns d for d or od for od with keys changed to lower case
"""
for k in d.iterkeys():
v = d.pop(k)
if (type(k) == str) and (not k.islower()):
k = k.lower()
d[k] = v
return d
It forms an infinite loop because of the way ordered dictionaries add new members (to the end)
Since you are using iterkeys, it is using a generator. When you assign d[k] = v you are adding the new key/value to the end of the dictionary. Because you are using a generator, that will continue to generate keys as you continue adding them.
You could fix this in a few ways. One would be to create a new ordered dict from the previous.
def key_lower(d):
newDict = OrderedDict()
for k, v in d.iteritems():
if (isinstance(k, (str, basestring))):
k = k.lower()
newDict[k] = v
return newDict
The other way would be to not use a generator and use keys instead of iterkeys
As sberry mentioned, the infinite loop is essentially as you are modifying and reading the dict at the same time.
Probably the simplest solution is to use OrderedDict.keys() instead of OrderedDict.iterkeys():
for k in d.keys():
v = d.pop(k)
if (type(k) == str) and (not k.islower()):
k = k.lower()
d[k] = v
as the keys are captured directly at the start, they won't get updated as items are changed in the dict.
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?
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.
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.
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.
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.
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
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()))
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
For python 3
dict((k, v) for k, v in metadata.items() if v)
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
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])}
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:
use isinstance instead of type
unroll the list comp into for loop for efficiency
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
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
"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'}
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
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
}
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_)
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
metadata ={'src':'1921','dest':'1337','email':'','movile':''}
ot = {k: v for k, v in metadata.items() if v != ''}
print(f"Final {ot}")
You also have an option with filter method:
filtered_metadata = dict( filter(lambda val: val[1] != u'', metadata.items()) )
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.
I'm trying to create a generic function that replaces dots in keys of a nested dictionary. I have a non-generic function that goes 3 levels deep, but there must be a way to do this generic. Any help is appreciated! My code so far:
output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}
def print_dict(d):
new = {}
for key,value in d.items():
new[key.replace(".", "-")] = {}
if isinstance(value, dict):
for key2, value2 in value.items():
new[key][key2] = {}
if isinstance(value2, dict):
for key3, value3 in value2.items():
new[key][key2][key3.replace(".", "-")] = value3
else:
new[key][key2.replace(".", "-")] = value2
else:
new[key] = value
return new
print print_dict(output)
UPDATE: to answer my own question, I made a solution using json object_hooks:
import json
def remove_dots(obj):
for key in obj.keys():
new_key = key.replace(".","-")
if new_key != key:
obj[new_key] = obj[key]
del obj[key]
return obj
output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}
new_json = json.loads(json.dumps(output), object_hook=remove_dots)
print new_json
Yes, there exists better way:
def print_dict(d):
new = {}
for k, v in d.iteritems():
if isinstance(v, dict):
v = print_dict(v)
new[k.replace('.', '-')] = v
return new
(Edit: It's recursion, more on Wikipedia.)
Actually all of the answers contain a mistake that may lead to wrong typing in the result.
I'd take the answer of #ngenain and improve it a bit below.
My solution will take care about the types derived from dict (OrderedDict, defaultdict, etc) and also about not only list, but set and tuple types.
I also do a simple type check in the beginning of the function for the most common types to reduce the comparisons count (may give a bit of speed in the large amounts of the data).
Works for Python 3. Replace obj.items() with obj.iteritems() for Py2.
def change_keys(obj, convert):
"""
Recursively goes through the dictionary obj and replaces keys with the convert function.
"""
if isinstance(obj, (str, int, float)):
return obj
if isinstance(obj, dict):
new = obj.__class__()
for k, v in obj.items():
new[convert(k)] = change_keys(v, convert)
elif isinstance(obj, (list, set, tuple)):
new = obj.__class__(change_keys(v, convert) for v in obj)
else:
return obj
return new
If I understand the needs right, most of users want to convert the keys to use them with mongoDB that does not allow dots in key names.
I used the code by #horejsek, but I adapted it to accept nested dictionaries with lists and a function that replaces the string.
I had a similar problem to solve: I wanted to replace keys in underscore lowercase convention for camel case convention and vice versa.
def change_dict_naming_convention(d, convert_function):
"""
Convert a nested dictionary from one convention to another.
Args:
d (dict): dictionary (nested or not) to be converted.
convert_function (func): function that takes the string in one convention and returns it in the other one.
Returns:
Dictionary with the new keys.
"""
new = {}
for k, v in d.iteritems():
new_v = v
if isinstance(v, dict):
new_v = change_dict_naming_convention(v, convert_function)
elif isinstance(v, list):
new_v = list()
for x in v:
new_v.append(change_dict_naming_convention(x, convert_function))
new[convert_function(k)] = new_v
return new
Here's a simple recursive solution that deals with nested lists and dictionnaries.
def change_keys(obj, convert):
"""
Recursivly goes through the dictionnary obj and replaces keys with the convert function.
"""
if isinstance(obj, dict):
new = {}
for k, v in obj.iteritems():
new[convert(k)] = change_keys(v, convert)
elif isinstance(obj, list):
new = []
for v in obj:
new.append(change_keys(v, convert))
else:
return obj
return new
You have to remove the original key, but you can't do it in the body of the loop because it will throw RunTimeError: dictionary changed size during iteration.
To solve this, iterate through a copy of the original object, but modify the original object:
def change_keys(obj):
new_obj = obj
for k in new_obj:
if hasattr(obj[k], '__getitem__'):
change_keys(obj[k])
if '.' in k:
obj[k.replace('.', '$')] = obj[k]
del obj[k]
>>> foo = {'foo': {'bar': {'baz.121': 1}}}
>>> change_keys(foo)
>>> foo
{'foo': {'bar': {'baz$121': 1}}}
You can dump everything to a JSON
replace through the whole string and load the JSON back
def nested_replace(data, old, new):
json_string = json.dumps(data)
replaced = json_string.replace(old, new)
fixed_json = json.loads(replaced)
return fixed_json
Or use a one-liner
def short_replace(data, old, new):
return json.loads(json.dumps(data).replace(old, new))
While jllopezpino's answer works but only limited to the start with the dictionary, here is mine that works with original variable is either list or dict.
def fix_camel_cases(data):
def convert(name):
# https://stackoverflow.com/questions/1175208/elegant-python-function-to-convert-camelcase-to-snake-case
s1 = re.sub('(.)([A-Z][a-z]+)', r'\1_\2', name)
return re.sub('([a-z0-9])([A-Z])', r'\1_\2', s1).lower()
if isinstance(data, dict):
new_dict = {}
for key, value in data.items():
value = fix_camel_cases(value)
snake_key = convert(key)
new_dict[snake_key] = value
return new_dict
if isinstance(data, list):
new_list = []
for value in data:
new_list.append(fix_camel_cases(value))
return new_list
return data
Here's a 1-liner variant of #horejsek 's answer using dict comprehension for those who prefer:
def print_dict(d):
return {k.replace('.', '-'): print_dict(v) for k, v in d.items()} if isinstance(d, dict) else d
I've only tested this in Python 2.7
I am guessing you have the same issue as I have, inserting dictionaries into a MongoDB collection, encountering exceptions when trying to insert dictionaries that have keys with dots (.) in them.
This solution is essentially the same as most other answers here, but it is slightly more compact, and perhaps less readable in that it uses a single statement and calls itself recursively. For Python 3.
def replace_keys(my_dict):
return { k.replace('.', '(dot)'): replace_keys(v) if type(v) == dict else v for k, v in my_dict.items() }