I have a dictionary, and I want to iterate over the keys, test the keys, and then print the keys that pass the test.
For example, my code looks like this at the moment:
x = {4:"a", 1:"b", 0:"c", 9:"d"}
for t in x:
if t > 2:
print t
However, all the output is, is just the key.
How do I get the code to print out the value at the end of the code, instead of the key?
Also, how do I get this in the form of a dictionary, {key : value}?
You could try this: print t, x[t]
Or this:
for key, val in x.items():
if key > 2:
print key, val
If you want to get only the pairs that you're printing into a new dictionary, then you have to put them there:
newdict = {}
for key, val in x.items():
if key > 2:
print key, val
# put all the keys > 2 into the new dict
newdict[key] = val
Of course that solution is printing when it inserts. If you're doing more than just a little script you will get more utility out of breaking functionality out, making sure that any given function does one, and only one, thing:
def filter_dict(x):
newdict = {}
for key, val in x.items():
if key > 2:
# put all the keys > 2 into the new dict
newdict[key] = val
def pretty_print(dct):
for k, v in dct.items():
print k, v
filtered = filter_dict(x)
pretty_print(dct)
This doesn't apply if you're just debugging, of course, and there are ways in which it depends on how large of a program you're writing: if you're doing a simple enough thing then the extra flexibility you get by breaking functionality up is lost because of the extra effort of figuring out exactly what any given thing does. That said: you should generally do it.
Additionally, the most idiomatic way to filter lists on a simple comparison in Python is to use a dict comprehension (new in Python 2.7/3.0):
filtered = {key: val for key, val in x if key > 2}
print [(k,v) for k,v in yourDict.items() if test(k)]
(You could just do k for k,v in... or v for k,v in.... If you only need values, you can use yourDict.values().)
x = {4:"a", 1:"b", 0:"c", 9:"d"}
for t in x:
if t > 2:
print '{}: {}'.format(t, x[t])
Slightly more pythonic:
>>> for key, value in x.iteritems():
... if key > 2:
... print '{}: {}'.format(key, value)
...
4: a
9: d
edit: To just print the value:
>>> for key, value in x.iteritems():
... if key > 2:
... print value
...
a
d
key as well as value:
x = {4:"a", 1:"b", 0:"c", 9:"d"}
for k,v in x.items():
if k>2: print"{%d : %s}"%(k,repr(v))
{4 : 'a'}
{9 : 'd'}
just value:
x = {4:"a", 1:"b", 0:"c", 9:"d"}
for k,v in x.items():
if k>2:print v
a
d
Just the keys:
>>> [k for k in x.iterkeys() if k > 2]
[4, 9]
Just the values:
>>> [v for k,v in x.iteritems() if k > 2]
['a', 'd']
Key-value pairs:
>>> [(k, v) for k,v in x.iteritems() if k > 2]
[(4, 'a'), (9, 'd')]
As a dict:
>>> dict((k,v) for k,v in x.iteritems() if k > 2)
{9: 'd', 4: 'a'}
If you want to create a dictionary with subset of items, you can also use a dictionary comprehension for Python 2.7+:
>>> x = {4:"a", 1:"b", 0:"c", 9:"d"}
>>> y = {k:v for k, v in x.iteritems() if k > 2}
>>> y
{9: 'd', 4: 'a'}
This is only a newer equivalent with only minor syntax differences for what Johnysweb shows in his answer:
>>> y = dict((k, v) for k, v in x.iteritems() if k > 2)
>>> y
{9: 'd', 4: 'a'}
The following list comprehension will print the key and value if key > 2.
Try this:
print [(key, value) for key, value in x.iteritems() if key > 2]
This solution prints the key t and the value in the dictionary x at key t.
x = {4:"a", 1:"b", 0:"c", 9:"d"}
for t in x:
if t > 2:
print t, x[t]
Using lambda function
(lambda x:[(k,v) for k,v in x.iteritems() if k > 2])(x)
Related
I have an existing list of Key, Value pairs in my current dictionary called total_list. I want to check my list to see if the length of each Key == 1 in total_list, I want to add that key and its value pair to a new dictionary. This is the code that I've come up with.
total_list = {104370544: [31203.7, 01234], 106813775: [187500.0], 106842625: [60349.8]}
diff_so = defaultdict(list)
for key, val in total_list:
if len(total_list[key]) == 1:
diff_so[key].append[val]
total_list.pop[key]
But I keep getting an error with
"cannot unpack non-iterable int object".
I was wondering if there's anyway for me to fix this code for it to run properly?
Assuming that the OP means a string of one character by length = 1 of the key.
You can do this:
total_list = [{'abc':"1", 'bg':"7", 'a':"7"}]
new_dict = {}
for i in total_list:
for k,v in i.items():
if len(k) == 1:
new_dict[str(k)] = v
else:
pass
print(new_dict)
Output:
{'a': '7'}
After edit:
total_list = {104370544: [31203.7, 1234], 106813775: [187500.0], 106842625: [60349.8]}
new_dict = {}
for k,v in total_list.items():
if len(v) == 1:
new_dict[k] = v
else:
pass
Output:
{'106842625': [60349.8], '106813775': [187500.0]}
You just need a dictionary comprehension
diff_so = {k: v for k, v in total_list.items() if len(v) == 1}
I have a dictionary contains lists of values and a list:
dict={'first':45, 'second':30, 'third':56}
list= [30,45]
I want to compare the value in the dictionary with the list and a match to add to a new dictionary after that, remove from the old dict all the values that are in the new dict: I'm doing something like this:
def get_sessions(self, talks):
result_sessions = {}
for k, v in self.sessions.items():
for i in talks:
if v == i:
result_sessions[k] = v
for k, v in result_sessions.items():
del self.sessions[k]
return result_sessions
Maybe you know a more elegant solution? any help?
This is one approach.
Ex:
d ={'first':45, 'second':30, 'third':56}
lst = [30,45]
result_sessions = {k: v for k, v in d.items() if v in lst}
d = { k : d[k] for k in set(d) - set(result_sessions) }
print(result_sessions)
print(d)
Output:
{'second': 30, 'first': 45}
{'third': 56}
I need to write a function which is doing following work
Find a string as value in a dictionary of dictionaries and return its key
(1st key if found in main dictionary, 2nd key if found in sub dictionary).
Source Code
Here is the function which I try to implement, but it works incorrect as I can't find any answer of how to convert list into dictionary as in this case the following error occurs
for v, k in l:
ValueError: need more than 1 value to unpack
def GetKeyFromDictByValue(self, dictionary, value_to_find):
""""""
key_list = [k for (k, v) in dictionary.items() if v == value_to_find]
if key_list.__len__() is not 0:
return key_list[0]
else:
l = [s for s in dictionary.values() if ":" in str(s)]
d = defaultdict(list)
for v, k in l:
d[k].append(v)
print d
dict = {'a': {'a1': 'a2'}, "aa": "aa1", 'aaa': {'aaa1': 'aaa2'}}
print GetKeyFromDictByValue(dict, "a2")
I must do this on Python 2.5
You created a list of only the dictionary values, but then try to loop over it as if it already contains both keys and values of those dictionaries. Perhaps you wanted to loop over each matched dictionary?
l = [v for v in dictionary.values() if ":" in str(v)]
d = defaultdict(list)
for subdict in l:
for k, v in subdict.items():
I'd instead flatten the structure:
def flatten(dictionary):
for key, value in dictionary.iteritems():
if isinstance(value, dict):
# recurse
for res in flatten(value):
yield res
else:
yield key, value
then just search:
def GetKeyFromDictByValue(self, dictionary, value_to_find):
for key, value in flatten(dictionary):
if value == value_to_find:
return key
Demo:
>>> sample = {'a': {'a1': 'a2'}, "aa": "aa1", 'aaa': {'aaa1': 'aaa2'}}
>>> GetKeyFromDictByValue(None, sample, "a2")
'a1'
Given a list of dictionaries ( each of which have same keys), I want total number of different values with which a given key is associated
$ li = [{1:2,2:3},{1:2,2:4}] $ the expected output is {1:1,2:2}
I came up with the following piece of code...Is there a better way of doing this ?
counts = {}
values = {}
for i in li:
for key,item in i.items():
try:
if item in values[key]:
continue
except KeyError:
else:
try:
counts[key] += 1
except KeyError:
counts[key] = 1
try:
values[key].append(item)
except KeyError:
values[key] = [item]
Something like this is probably more direct:
from collections import defaultdict
counts = defaultdict(set)
for mydict in li:
for k, v in mydict.items():
counts[k].add(v)
That takes care of the collecting / counting of the values. To display them like you want them, this would get you there:
print dict((k, len(v)) for k, v in counts.items())
# prints {1: 1, 2: 2}
Here is yet another alternative:
from collections import defaultdict
counts = defaultdict(int)
for k, v in set(pair for d in li for pair in d.items()):
counts[k] += 1
And the result:
>>> counts
defaultdict(<type 'int'>, {1: 1, 2: 2})
You could so something like this:
li = [{1:2,2:3},{1:2,2:4}]
def makesets(x, y):
for k, v in x.iteritems():
v.add(y[k])
return x
distinctValues = reduce(makesets, li, dict((k, set()) for k in li[0].keys()))
counts = dict((k, len(v)) for k, v in distinctValues.iteritems())
print counts
When I run this it prints:
{1: 1, 2: 2}
which is the desired result.
counts = {}
values = {}
for i in li:
for key,item in i.items():
if not (key in values.keys()):
values[key] = set()
values[key].add(item)
for key in values.keys():
counts[key] = len(values[key])
using flattening list in case dicts are not alway same length:
li=[{1: 2, 2: 3}, {1: 2, 2: 4}, {1: 3}]
dic={}
for i,j in [item for sublist in li for item in sublist.items()]:
dic[i] = dic[i]+1 if i in dic else 1
d = {'foo': 'x',
'bar': 'y',
'zoo': 'None',
'foobar': 'None'}
I want to filter all the items whose value is 'None' and update the foo and bar items with a particular value. I tried:
for i in x.items():
....: if i[i] == 'None':
....: x.pop(i[0])
....: else:
....: x.update({i[0]:'updated'})
But it is not working.
It is not clear what is 'None' in the dictionary you posted. If it is a string, you can use the following:
dict((k, 'updated') for k, v in d.items() if v != 'None')
If it is None, just replace the checking, for example:
dict((k, 'updated') for k, v in d.items() if v is None)
(If you are still using Python 2, replace .items() with .iteritems())
it's not clear where you're getting your 'updated' value from, but in general it would look like this:
{i: 'updated' for i, j in d.items() if j != 'None'}
in python2.7 or newer.
Something like this should work
>>> for x in [x for x in d.keys() if d[x] == 'None']:
d.pop(x)
You could try writing a general filter function:
def filter(dict, by_key = lambda x: True, by_value = lambda x: True):
for k, v in dict.items():
if (by_key(k) and by_value(v)):
yield (k, v)
or
def filter(dict, by_key = lambda x: True, by_value = lambda x: True):
return dict((k, v) for k, v in dict.items() if by_key(k) and by_value(v))
and then you can do this:
new_d = dict(filter(d, by_key = lambda k: k != 'None'))
Note that this was written to be compatible with Python 3. Mileage may vary.
The cool thing about this is that it allows you to pass arbitrary lambdas in to filter instead of hard-coding a particular condition. Plus, it's not any slower than any of the other solutions here. Furthermore, if you use the first solution I provided then you can iterate over the results.
new_d = dict((k, 'updated') for k, v in d.iteritems() if k != 'None')