So I have a P = dict(). I have the following code:
def someFunction():
P['key'] += 1
'''do other task'''
What is the simplest way to check if P['key'] is defined or not?
I checked How do I check if a variable exists? but I am not sure if that answers my question.
Two main ways to check in an ordinary dict:
The "look before you leap" paradigm. The else statement isn't required, of course, unless you want to define some alternate behavior:
if 'key' in P:
P['key'] += 1
else:
pass
The "easier to ask for forgiveness than permission" paradigm:
try:
P['key'] += 1
except KeyError:
pass # Or do something else
Or you could use a defaultdict as suggested.
You should use a defaultdict from the collections module.
from collections import defaultdict
d = defaultdict(int)
d[0] = 5
d[1] = 10
for i in range(3):
d[i] += 1
# Note that d[2] was not set before the loop
for k, v in d.items():
print('%i: %i' % (k,v))
prints:
brunsgaard#archbook /tmp> python test.py
0: 6
1: 11
2: 1
Usually I will check key presence with
if some_key in some_dict:
print("do something")
Advanced usage: If you have a dictionary, key is string, value is a list. When key exists, you want to add an element to key associated value list. So you can
some_dict[some_key] = some_dict.get(some_key, []) + [new_item];
Related
How can I check that my dict contains only one value filled ?
I want to enter in my condition only if the value is the only one in my dict and of this type (in my example "test2") of my dict.
For now I have this if statement
my_dict = {}
my_dict["test1"] = ""
my_dict["test2"] = "example"
my_dict["test3"] = ""
my_dict["test4"] = ""
if my_dict["test2"] and not my_dict["test1"] and not my_dict["test3"] and not my_dict["test4"]:
print("inside")
I would like to find a better, classy and "pep8" way to achieve that
Any ideas ?
You have to check every value for truthiness, there's no way around that, e.g.
if sum(1 for v in my_dict.values() if v) == 1:
print('inside')
You can use filter() as below to check how many values are there in the dictionary.
if len(list(filter(None, my_dict.values()))) == 1:
print("inside")
Assuming that all your values are strings, what about
ref_key = "test2"
if ''.join(my_dict.values()) == my_dict[ref_key]:
print("inside")
... since it looks like you have a precise key in mind (when you do if my_dict["test2"]). Otherwise, my answer is (twice) less general than (some) others'.
Maybe you want to check if there's only one pair in dictionary after removing the empty values.
my_dict = {}
my_dict["test1"] = ""
my_dict["test2"] = "example"
my_dict["test3"] = ""
my_dict["test4"] = ""
my_dict={key:val for key,val in my_dict.items() if val}
if len(my_dict)==1:
print("inside")
Here is the another flavour (without loops):
data = list(my_dict.values())
if data.count('') + 1 == len(data):
print("Inside")
This is relatively straightforward but more pointed question on how to write the most concise if/any/for loop that accomplishes what I need. I'm rather new to Python, and much more used to the C++ way of doing things.
I need to do something like the following:
for item in my_dict:
if some_var == item.some_attribute:
if item.another_attribute exists:
variable = "A"
else variable = "B"
print "Duplicate item exists in my list of type: {}".format(variable)
I've condensed it into something like this:
if any(some_var == item.some_attribute for item in my_dict):
variable = "A" if item.another_attribute else "B"
print "Duplicate item exists in my list of type: {}".format(variable)
However, when doing this, I'm getting an "unresolved reference" error for item. Any ideas as to how I can write a concise loop, equivalence check, and presence check like I've described above in a way that allows me to access and perform methods on item?
Thank you!
EDIT: thank you so much for the answer #pjhaugh, that's exactly what I was looking for. Works beautifully! Thank you everyone for your helpful comments as well.
What's happening here is that item does not leak out of the generator expression scope (this is a good thing). You could obtain item by getting the next thing yielded by a generator, or catching that generator yielding nothing.
try:
item = next(item for item in my_dict if some_var == item.some_attribute)
variable = "A" if item.another_attribute else "B"
except StopIteration:
# there is no such item in my_dict
You can extract the objects yielded by an iterator in a for loop like this
for item in (i for i in my_dict if some_var == item.some_attribute):
variable = "A" if item.another_attribute is not None else "B"
print("Duplicate item exists in my list of type: {}".format(variable))
break
else: # no break encountered
... # not found
Currently you're iterating over the keys of the dictionary. If you want the items (pairs in c++ lingo), you need my_dict.items().
As pointed out in my comment - the any(some_var == item.some_attribute for item in my_dict) builds its own local scope. You cannot use item outside of it.
You can however loop over your dict like so:
Create minimal verifyable complete example data:
class K:
def __init__(self):
self.some_attribute = None
self.another_attribute = None
def __repr__(self):
return "{} - {}".format(self.some_attribute,self.another_attribute)
def __str__(self):
return "{} - {}".format(self.some_attribute,self.another_attribute)
k1 = K()
k1.some_attribute = 99
k2 = K()
k2.some_attribute = 33
k3 = K()
k3.some_attribute = 33
k3.another_attribute = 1
my_dict = { 1: k1, 2:k2, 3:k3}
some_var = 33
You can use a for loop like so:
variable = None
for item in my_dict.items():
k,v = item
if v.some_attribute != some_var:
print "No dupe: {}".format(v)
continue
if v.some_attribute == some_var and v.another_attribute is not None:
variable = "A"
else:
variable = "B"
print "Duplicate item exists in my list of type: {}".format(variable)
print v
Output:
No dupe: 99 - None
Duplicate item exists in my list of type: B
33 - None
Duplicate item exists in my list of type: A
33 - 1
I'm trying to make a context-free grammar simplification software.
I'm stuck when it comes to delete some specific items from the dictionary's values or even the key value pair.
The problem is that it doesn't follow a pattern.
If the element belongs to V1, I need to keep it in dictionary.
(V1 is the list of all values who derivates a terminal, those guys are the only ones I need to keep on my dictionary, but it's not that simple)
If the element doesn't belongs to V1 and dictionary's values is a string, I need to remove the element.
If the element doesn't belongs to V1 and dictionary's values is a list, I need to check if it's the single element of that list, if so, delete Value.
The failed loop is down here.
I printed the parts that I can't figure out the logic in modifying the dictionary.
counter = 0
for k,v in derivations.items():
derivationsCount = len(v)
while counter < derivationsCount:
if lista_ou_string(v[counter]): # returns True for lists, False for else
sizeOfList = len(v[counter])
counter2 = 0
while counter2 <= (sizeOfList - 1):
if v[counter][counter2] not in V1:
if derivationsCount == 1:
print("# NEED TO DELETE BOTH KEY AND VALUE FROM derivatios.items()")
else:
print("# NEED TO DELETE ONLY THE VALUE FROM derivations.items()")
counter2 += 1
else: # strings \/
if v[counter] not in V1:
if derivationsCount == 1:
print("# NEED TO DELETE BOTH KEY AND VALUE FROM derivatios.items()")
else:
print("# NEED TO DELETE ONLY THE VALUE FROM derivations.items()")
else:
print("# DO NOT DELETE ANYTHING! ALL LISTS ELEMENTS BELONGS TO 'V1'")
counter += 1
One does not want to modify a dictionary (or list) while looping over it. Therefore I create a copy of the derivations - new_derivations and modify this new_derivations:
import copy
new_derivations = copy.deepcopy(derivations)
for k, v in derivations.items():
for vi in v:
if (lista_ou_string(vi) and not set(vi).issubset(V1)) or vi not in V1:
if len(v) == 1:
# NEED TO DELETE BOTH KEY AND VALUE FROM derivatios.items()
del new_derivations[k]
break
else:
# NEED TO DELETE ONLY THE VALUE FROM derivations.items()
idx = new_derivations[k].index(vi)
del new_derivations[k][idx]
I would actually implement the above code differently: instead of thinking in terms of removing items from derivations, think instead of when an element should be added to the list. Then the code becomes much simpler:
new_derivations = {}
for k, v in derivations.items():
nv = [vi for vi in v if ((isinstance(vi, list) and set(vi).issubset(V1))
or vi in V1)]
if nv:
new_derivations[k] = nv
if you want to delete a key,value pair from a dictionary, use del:
>>> my_dictionary = {'foo':'bar', 'boo':'baz'}
>>> del my_dictionary['foo']
>>> my_dictionary
{'boo': 'baz'}
if you want to delete the value, but keep the key, you can try assigning key None:
>>> my_dictionary = {'foo':'bar', 'boo':'baz'}
>>> my_dictionary['foo'] = None
>>> my_dictionary
{'foo': None, 'boo': 'baz'}
I have code that works but I'm wondering if there is a more pythonic way to do this. I have a dictionary and I want to see if:
a key exists
that value isn't None (NULL from SQL in this case)
that value isn't simply quote quote (blank?)
that value doesn't solely consist of spaces
So in my code the keys of "a", "b", and "c" would succeed, which is correct.
import re
mydict = {
"a":"alpha",
"b":0,
"c":False,
"d":None,
"e":"",
"g":" ",
}
#a,b,c should succeed
for k in mydict.keys():
if k in mydict and mydict[k] is not None and not re.search("^\s*$", str(mydict[k])):
print(k)
else:
print("I am incomplete and sad")
What I have above works, but that seems like an awfully long set of conditions. Maybe this simply is the right solution but I'm wondering if there is a more pythonic "exists and has stuff" or better way to do this?
UPDATE
Thank you all for wonderful answers and thoughtful comments. With some of the points and tips, I've updated the question a little bit as there some conditions I didn't have which should also succeed. I have also changed the example to a loop (just easier to test right?).
Try to fetch the value and store it in a variable, then use object "truthyness" to go further on with the value
v = mydict.get("a")
if v and v.strip():
if "a" is not in the dict, get returns None and fails the first condition
if "a" is in the dict but yields None or empty string, test fails, if "a" yields a blank string, strip() returns falsy string and it fails too.
let's test this:
for k in "abcde":
v = mydict.get(k)
if v and v.strip():
print(k,"I am here and have stuff")
else:
print(k,"I am incomplete and sad")
results:
a I am here and have stuff
b I am incomplete and sad # key isn't in dict
c I am incomplete and sad # c is None
d I am incomplete and sad # d is empty string
e I am incomplete and sad # e is only blanks
if your values can contain False, 0 or other "falsy" non-strings, you'll have to test for string, in that case replace:
if v and v.strip():
by
if v is not None and (not isinstance(v,str) or v.strip()):
so condition matches if not None and either not a string (everything matches) or if a string, the string isn't blank.
The get method for checking if a key exists is more efficient that iterating through the keys. It checks to see if the key exists without iteration using an O(1) complexity as apposed to O(n). My preferred method would look something like this:
if mydict.get("a") is not None and str(mydict.get("a")).replace(" ", "") != '':
# Do some work
You can use a list comprehension with str.strip to account for whitespace in strings.
Using if v is natural in Python to cover False-like objects, e.g. None, False, 0, etc. So note this only works if 0 is not an acceptable value.
res = [k for k, v in mydict.items() if (v.strip() if isinstance(v, str) else v)]
['a']
Here's a simple one-liner to check:
The key exists
The key is not None
The key is not ""
bool(myDict.get("some_key"))
As for checking if the value contains only spaces, you would need to be more careful as None doesn't have a strip() method.
Something like this as an example:
try:
exists = bool(myDict.get('some_key').strip())
except AttributeError:
exists = False
Well I have 2 suggestions to offer you, especially if your main issue is the length of the conditions.
The first one is for the check if the key is in the dict. You don't need to use "a" in mydict.keys() you can just use "a" in mydict.
The second suggestion to make the condition smaller is to break down into smaller conditions stored as booleans, and check these in your final condition:
import re
mydict = {
"a":"alpha",
"c":None,
"d":"",
"e":" ",
}
inKeys = True if "a" in mydict else False
isNotNone = True if mydict["a"] is not None else False
isValidKey = True if not re.search("^\s*$", mydict["a"]) else False
if inKeys and isNotNone and isValidKey:
print("I am here and have stuff")
else:
print("I am incomplete and sad")
it check exactly for NoneType not only None
from types import NoneType # dont forget to import this
mydict = {
"a":"alpha",
"b":0,
"c":False,
"d":None,
"e":"",
"g":" ",
}
#a,b,c should succeed
for k in mydict:
if type(mydict[k]) != NoneType:
if type(mydict[k]) != str or type(mydict[k]) == str and mydict[k].strip():
print(k)
else:
print("I am incomplete and sad")
else:
print("I am incomplete and sad")
cond is a generator function responsible for generating conditions to apply in a short-circuiting manner using the all function. Given d = cond(), next(d) will check if a exists in the dict, and so on until there is no condition to apply, in that case all(d) will evaluate to True.
mydict = {
"a":"alpha",
"c":None,
"d":"",
"e":" ",
}
def cond ():
yield 'a' in mydict
yield mydict ['a']
yield mydict ['a'].strip ()
if all (cond ()):
print("I am here and have stuff")
else:
print("I am incomplete and sad")
I am trying to find corresponding keys in two different dictionaries. Each has about 600k entries.
Say for example:
myRDP = { 'Actinobacter': 'GATCGA...TCA', 'subtilus sp.': 'ATCGATT...ACT' }
myNames = { 'Actinobacter': '8924342' }
I want to print out the value for Actinobacter (8924342) since it matches a value in myRDP.
The following code works, but is very slow:
for key in myRDP:
for jey in myNames:
if key == jey:
print key, myNames[key]
I've tried the following but it always results in a KeyError:
for key in myRDP:
print myNames[key]
Is there perhaps a function implemented in C for doing this? I've googled around but nothing seems to work.
Thanks.
Use sets, because they have a built-in intersection method which ought to be quick:
myRDP = { 'Actinobacter': 'GATCGA...TCA', 'subtilus sp.': 'ATCGATT...ACT' }
myNames = { 'Actinobacter': '8924342' }
rdpSet = set(myRDP)
namesSet = set(myNames)
for name in rdpSet.intersection(namesSet):
print name, myNames[name]
# Prints: Actinobacter 8924342
You could do this:
for key in myRDP:
if key in myNames:
print key, myNames[key]
Your first attempt was slow because you were comparing every key in myRDP with every key in myNames. In algorithmic jargon, if myRDP has n elements and myNames has m elements, then that algorithm would take O(n×m) operations. For 600k elements each, this is 360,000,000,000 comparisons!
But testing whether a particular element is a key of a dictionary is fast -- in fact, this is one of the defining characteristics of dictionaries. In algorithmic terms, the key in dict test is O(1), or constant-time. So my algorithm will take O(n) time, which is one 600,000th of the time.
in python 3 you can just do
myNames.keys() & myRDP.keys()
for key in myRDP:
name = myNames.get(key, None)
if name:
print key, name
dict.get returns the default value you give it (in this case, None) if the key doesn't exist.
You could start by finding the common keys and then iterating over them. Set operations should be fast because they are implemented in C, at least in modern versions of Python.
common_keys = set(myRDP).intersection(myNames)
for key in common_keys:
print key, myNames[key]
Best and easiest way would be simply perform common set operations(Python 3).
a = {"a": 1, "b":2, "c":3, "d":4}
b = {"t1": 1, "b":2, "e":5, "c":3}
res = a.items() & b.items() # {('b', 2), ('c', 3)} For common Key and Value
res = {i[0]:i[1] for i in res} # In dict format
common_keys = a.keys() & b.keys() # {'b', 'c'}
Cheers!
Use the get method instead:
for key in myRDP:
value = myNames.get(key)
if value != None:
print key, "=", value
You can simply write this code and it will save the common key in a list.
common = [i for i in myRDP.keys() if i in myNames.keys()]
Copy both dictionaries into one dictionary/array. This makes sense as you have 1:1 related values. Then you need only one search, no comparison loop, and can access the related value directly.
Example Resulting Dictionary/Array:
[Name][Value1][Value2]
[Actinobacter][GATCGA...TCA][8924342]
[XYZbacter][BCABCA...ABC][43594344]
...
Here is my code for doing intersections, unions, differences, and other set operations on dictionaries:
class DictDiffer(object):
"""
Calculate the difference between two dictionaries as:
(1) items added
(2) items removed
(3) keys same in both but changed values
(4) keys same in both and unchanged values
"""
def __init__(self, current_dict, past_dict):
self.current_dict, self.past_dict = current_dict, past_dict
self.set_current, self.set_past = set(current_dict.keys()), set(past_dict.keys())
self.intersect = self.set_current.intersection(self.set_past)
def added(self):
return self.set_current - self.intersect
def removed(self):
return self.set_past - self.intersect
def changed(self):
return set(o for o in self.intersect if self.past_dict[o] != self.current_dict[o])
def unchanged(self):
return set(o for o in self.intersect if self.past_dict[o] == self.current_dict[o])
if __name__ == '__main__':
import unittest
class TestDictDifferNoChanged(unittest.TestCase):
def setUp(self):
self.past = dict((k, 2*k) for k in range(5))
self.current = dict((k, 2*k) for k in range(3,8))
self.d = DictDiffer(self.current, self.past)
def testAdded(self):
self.assertEqual(self.d.added(), set((5,6,7)))
def testRemoved(self):
self.assertEqual(self.d.removed(), set((0,1,2)))
def testChanged(self):
self.assertEqual(self.d.changed(), set())
def testUnchanged(self):
self.assertEqual(self.d.unchanged(), set((3,4)))
class TestDictDifferNoCUnchanged(unittest.TestCase):
def setUp(self):
self.past = dict((k, 2*k) for k in range(5))
self.current = dict((k, 2*k+1) for k in range(3,8))
self.d = DictDiffer(self.current, self.past)
def testAdded(self):
self.assertEqual(self.d.added(), set((5,6,7)))
def testRemoved(self):
self.assertEqual(self.d.removed(), set((0,1,2)))
def testChanged(self):
self.assertEqual(self.d.changed(), set((3,4)))
def testUnchanged(self):
self.assertEqual(self.d.unchanged(), set())
unittest.main()
def combine_two_json(json_request, json_request2):
intersect = {}
for item in json_request.keys():
if item in json_request2.keys():
intersect[item]=json_request2.get(item)
return intersect