list1 = [99,55]
dict1 = {'k1':[],'k2':[]}
for num in list1:
if num > 77:
dict1['k1'].append(num)
else:
dict1['k2'].append(num)
print(dict1)
{'k1':[99],'k2':[55]}
But when I replaced dict1 = {'k1':[],'k2':[]} to {}.fromkeys(['k1','k2'],[]) , the result became {'k1': [99, 55], 'k2': [99, 55]}
why this happens? I really have no idea.
This happens because you are passing the same list object to both keys. This is the same situation as when you create an alias for a variable:
a = []
b = a
a.append(55)
b.append(99)
print(b)
prints [55, 99] because it is the same list instance.
If you want to make it more concise from a list of keys to initialize with empty list, you can do this:
dict1 = {k: [] for k in ('k1', 'k2')}
This will create a new list instance for every key.
Alternatively, you can use defaultdict
from collections import defaultdict
list1 = [99,55]
dict1 = defaultdict(list)
for num in list1:
if num > 77:
dict1['k1'].append(num)
else:
dict1['k2'].append(num)
print(dict1)
Also works.
The fromKeys() can also be supplied with a mutable object as the default value.
if we append value in the original list, the append takes place in all the values of keys.
example:
list1 = ['a', 'b', 'c', 'd']
list2 = ['SALIO']
dict1 = dict.fromkeys(list1, list2)
print(dict1)
output:
{'a': ['SALIO'], 'b': ['SALIO'], 'c': ['SALIO'], 'd': ['SALIO']}
then you can use this:
list1 = ['k1', 'k2']
dict1 = {'k1':[],'k2':[]}
list2 =[99,55]
for num in list2:
if num > 77:
a = ['k1']
dict1 = dict.fromkeys(a, [num])
else:
b = ['k2']
dict2 = dict.fromkeys(b,[num] )
res = {**dict1, **dict2}
print(res)
output:
{'k1': [99], 'k2': [55]}
You can also use the python code to merge dict code:
this function:
def Merge(dict1, dict2):
return(dict2.update(dict1))
then:
print(Merge(dict1, dict2)) #This return None
print(dict2) # changes made in dict2
Related
Suppose I have a 2 lists in my python script:
my_list = ['hat', 'bat']
other_list = ['A', 'B', 'C']
I want to iterate through other_list and create a nested list for 'bat's that adds '_' + other_list item to a the 'bat' and puts it in a nested list:
for item in other_list:
for thing in my_list:
if thing == 'bat':
print(thing + '_' + item)
My desired outcome would be new_list = ['hat',['bat_A', 'bat_B', 'bat_C']]
How would I achieve this?
I tried the below, but it produces this: ['hat', 'hat', 'hat', ['bat_A', 'bat_B', 'bat_C']]
new_list = []
extra = []
for item in other_list:
for thing in my_list:
if thing == 'bat':
extra.append(thing + '_' + item)
else:
new_list.append(thing)
new_list.append(extra)
Try this:
>>> my_list = ['hat', 'bat']
>>> other_list = ['A', 'B', 'C']
>>> new_list=[my_list[0], [f'{my_list[1]}_{e}' for e in other_list]]
>>> new_list
['hat', ['bat_A', 'bat_B', 'bat_C']]
If your question (which is a little unclear) is just about reacting to 'bat' with a different reaction, you can do this:
my_list = ['hat', 'bat','cat']
other_list = ['A', 'B', 'C']
new_list=[]
for e in my_list:
if e=='bat':
new_list.append([f'{e}_{x}' for x in other_list])
else:
new_list.append(e)
>>> new_list
['hat', ['bat_A', 'bat_B', 'bat_C'], 'cat']
Which can be reduced to:
>>> [[f'{e}_{x}' for x in other_list] if e=='bat' else e for e in my_list]
['hat', ['bat_A', 'bat_B', 'bat_C'], 'cat']
I think will work
my_list = ['hat', 'bat']
other = ['A', 'B' , 'C']
new_list = []
extra = []
for item in my_list:
if item == 'bat':
for char in other:
extra.append(item + '_' + char)
else:
new_list.append(item)
new_list.append(extra)
print(new_list)
OK, this is just my answer, but it seems to work. I think is clunky though, and I'm hoping for a better answer
my_list = ['hat', 'bat']
other_list = ['A', 'B', 'C']
new_list = []
extra = []
for item in other_list:
for thing in my_list:
if thing == 'bat':
extra.append(thing + '_' + item)
else:
if thing not in new_list:
new_list.append(thing)
new_list.append(extra)
python3
>>> a = dict()
>>> a['id1'] = dict()
>>> a['id1']['a'] = 5
>>> a['id1'] = dict()
>>> a['id1']['b'] = 10
>>> a
{'id1': {'b': 10}}
>>>
How can I check the existence of a['id1'] if dict or not and only if not then do a['id1'] = dict()
I need to print
{'id1': {'a': 5, 'b': 10}}
In PHP we don't need to define an associate array, we can assign it directly.
You can test for the presence of key k in a with:
if 'k' in a:
You can see if some object x is a dict with:
if isinstance(x, dict):
Use defaultdict:
from collections import defaultdict
a = defaultdict(dict)
a['id1']['a'] = 5
a['id1']['b'] = 10
Basically, I am trying to extract the values from one dictionary and update the value in another dictionary. I have four lists as follows:
a = [1,1,2,3,4,5]
b = [0,3,0,5,6,0]
c = [2,3,4,5,6,5]
d = [20,30,40,50,60,70]
So I use a defaultdict to store key,value pairs for a,b like:
one = defaultdict(list)
for k, v in zip(a, b):
one[k].append(v)
two = defaultdict(list)
for k, v in zip(c, d):
two[k].append(v)
Essentially, b is linked to c so I am trying to extract the values in the two dictionary and then update
the values in the one dictionary
So in the end one would look like {1: 30, 3: 50, 4: 60}
This is my code:
three = defaultdict(list)
for k, v in one.items():
if v in two.keys():
newvalue = two[v].values()
three[k].append(newvalue)
But I am now getting an error at line if v in two.keys(): as unhashable type: 'list'. I'm so lost, all
I want to do is use the values from one dictionary and then use those values to find the keys (which are the values
from the other table) and then get those corressponding values.
You are creating a dictionary of list in the beginning:
one = defaultdict(list)
for k, v in zip(a, b):
one[k].append(v)
[output] : defaultdict(list, {1: [0, 3], 2: [0], 3: [5], 4: [6], 5: [0]})
two = defaultdict(list)
for k, v in zip(c, d):
two[k].append(v)
[output] : defaultdict(list, {2: [20], 3: [30], 4: [40], 5: [50, 70], 6: [60]})
Therefore when calling k,v in one.items(), you are getting a key and a list.
Simply switch to iterate through the list , and you should be good to go
three = defaultdict(list)
for k, v in one.items():
for value in v:
if value in two.keys():
newvalue = two[value]
three[k].append(newvalue)
However I'm getting this output :
defaultdict(list, {1: [[30]], 3: [[50, 70]], 4: [[60]]})
Which sounds reasonable to me, but it is not your expected one, can you please explain ?
Let's try know with dic comprehension
output = { k : two[v_2] for k,v in one.items() for v_2 in v}
[output] : {1: [30], 2: [], 3: [50, 70], 4: [60], 5: []}
Request to sum :
Of course, multiple ways of doing it , the quickest is again with dict_comprehension and sum
output_sum = {k: sum(v) for k,v in output.items()}
I am trying to write a function to extract only words unique to each key and list them in a dictionary output like {"key1": "unique words", "key2": "unique words", ... }. I start out with a dictionary. To test with I created a simple dictionary:
d = {1:["one", "two", "three"], 2:["two", "four",
"five"], 3:["one","four", "six"]}
My output should be:
{1:"three",
2:"five",
3:"six"}
I am thinking maybe split in to separate lists
def return_unique(dct):
Klist = list(dct.keys())
Vlist = list(dct.values())
aList = []
for i in range(len(Vlist)):
for j in Vlist[i]:
if
What I'm stuck on is how do I tell Python to do this: if Vlist[i][j] is not in the rest of Vlist then aList.append(Vlist[i][j]).
Thank you.
You can try something like this:
def return_unique(data):
all_values = []
for i in data.values(): # Get all values
all_values = all_values + i
unique_values = set([x for x in all_values if all_values.count(x) == 1]) # Values which are not duplicated
for key, value in data.items(): # For Python 3.x ( For Python 2.x -> data.iteritems())
for item in value: # Comparing values of two lists
for item1 in unique_values:
if item == item1:
data[key] = item
return data
d = {1:["one", "two", "three"], 2:["two", "four", "five"], 3:["one","four", "six"]}
print (return_unique(d))
result >> {1: 'three', 2: 'five', 3: 'six'}
Since a key may have more than one unique word associated with it, it makes sense for the values in the new dictionary to be a container type object to hold the unique words.
The set difference operator returns the difference between 2 sets:
>>> a = set([1, 2, 3])
>>> b = set([2, 4, 6])
>>> a - b
{1, 3}
We can use this to get the values unique to each key. Packaging these into a simple function yields:
def unique_words_dict(data):
res = {}
values = []
for k in data:
for g in data:
if g != k:
values += data[g]
res[k] = set(data[k]) - set(values)
values = []
return res
>>> d = {1:["one", "two", "three"],
2:["two", "four", "five"],
3:["one","four", "six"]}
>>> unique_words_dict(d)
{1: {'three'}, 2: {'five'}, 3: {'six'}}
If you only had to do this once, then you might be interested in the less efficeint but more consice dictionary comprehension:
>>> from functools import reduce
>>> {k: set(d[k]) - set(reduce(lambda a, b: a+b, [d[g] for g in d if g!=k], [])) for k in d}
{1: {'three'}, 2: {'five'}, 3: {'six'}}
I have two python dicts:
payload = {"key1":{"a":"1"},"key2":{"b":"2","c":"3"}}
and
data = {"1":"John","2":"Jacob"}
I would like my output to be:
{"key1":{"a":"John"},"key2":{"b":"Jacob","c":""}}
Any method that I try correctly prints the values, but does not update the output dictionary.
You can do something like this using dict comprehension :
payload = {"key1":{"a":"1"},"key2":{"b":"2","c":"3"}}
data = {"1":"John","2":"Jacob"}
final = {k: {i:data[j] if j in data.keys() else "" for i, j in payload[k].items()} for k in payload}
print(final)
Output:
{'key2': {'b': 'Jacob', 'c': ''}, 'key1': {'a': 'John'}}
There is no single method for this I am aware of, but you can use:
for k, v in payload.viewitems():
payload[k] = {}
for kv, vv in v.viewitems():
payload[k][kv] = data.get(vv, "")
if you then inspect payload it has the contents you are after:
{'key2': {'c': '', 'b': 'Jacob'}, 'key1': {'a': 'John'}}