I wonder whether there's a quicker and less time consuming way to iterate over a list of tuples, finding the right match. What I do is:
# this is a very long list.
my_list = [ (old1, new1), (old2, new2), (old3, new3), ... (oldN, newN)]
# go through entire list and look for match
for j in my_list:
if j[0] == VALUE:
PAIR_FOUND = True
MATCHING_VALUE = j[1]
break
this code can take quite some time to execute, depending on the number of items in the list. I'm sure there's a better way of doing this.
I think that you can use
for j,k in my_list:
[ ... stuff ... ]
Assuming a bit more memory usage is not a problem and if the first item of your tuple is hashable, you can create a dict out of your list of tuples and then looking up the value is as simple as looking up a key from the dict. Something like:
dct = dict(tuples)
val = dct.get(key) # None if item not found else the corresponding value
EDIT: To create a reverse mapping, use something like:
revDct = dict((val, key) for (key, val) in tuples)
The question is dead but still knowing one more way doesn't hurt:
my_list = [ (old1, new1), (old2, new2), (old3, new3), ... (oldN, newN)]
for first,*args in my_list:
if first == Value:
PAIR_FOUND = True
MATCHING_VALUE = args
break
The code can be cleaned up, but if you are using a list to store your tuples, any such lookup will be O(N).
If lookup speed is important, you should use a dict to store your tuples. The key should be the 0th element of your tuples, since that's what you're searching on. You can easily create a dict from your list:
my_dict = dict(my_list)
Then, (VALUE, my_dict[VALUE]) will give you your matching tuple (assuming VALUE exists).
I wonder whether the below method is what you want.
You can use defaultdict.
>>> from collections import defaultdict
>>> s = [('red',1), ('blue',2), ('red',3), ('blue',4), ('red',1), ('blue',4)]
>>> d = defaultdict(list)
>>> for k, v in s:
d[k].append(v)
>>> sorted(d.items())
[('blue', [2, 4, 4]), ('red', [1, 3, 1])]
Related
I have an OrderedDict in Python, and I only want to get the first key-vale pairs. How to get it? For example, to get the first 4 elements, i did the following:
subdict = {}
for index, pair in enumerate(my_ordered_dict.items()):
if index < 4:
subdict[pair[0]] = pair[1]
Is this the good way to do it?
That approach involves running over the whole dictionary even though you only need the first four elements, checking the index over and over, and manually unpacking the pairs, and manually performing index checking unnecessarily.
Making it short-circuit is easy:
subdict = {}
for index, pair in enumerate(my_ordered_dict.items()):
if index >= 4:
break # Ends the loop without iterating all of my_ordered_dict
subdict[pair[0]] = pair[1]
and you can nested the unpacking to get nicer names:
subdict = {}
# Inner parentheses mandatory for nested unpacking
for index, (key, val) in enumerate(my_ordered_dict.items()):
if index >= 4:
break # Ends the loop
subdict[key] = value
but you can improve on that with itertools.islice to remove the manual index checking:
from itertools import islice # At top of file
subdict = {}
# islice lazily produces the first four pairs then stops for you
for key, val in islice(my_ordered_dict.items(), 4):
subdict[key] = value
at which point you can actually one-line the whole thing (because now you have an iterable of exactly the four pairs you want, and the dict constructor accepts an iterable of pairs):
subdict = dict(islice(my_ordered_dict.items(), 4))
You can use a map function, like this
item = dict(map(lambda x: (x, subdict[x]),[*subdict][:4]))
Here is one approach:
sub_dict = dict(pair for i, pair in zip(range(4), my_ordered_dict.items()))
The length of zip(a,b) is equal to the length of the shortest of a and b, so if my_ordered_dict.items() is longer than 4, zip(range(4), my_ordered_dict.items() just takes the first 4 items. These key-value pairs are passed to the dict builtin to make a new dict.
If I have list of integers and a function getErrorType(int) that returns some enum type, what's a Pythonic way to get a dictionary where the key is the enum type and value is the count of how many values in the array returned that error type?
Example:
arr = [1, 2, 3]
getErrorType(1) returns EXCEPTION
getErrorType(2) returns MALFORMED_DATA
getErrorType(3) returns EXCEPTION
I want to be able to get: {EXCEPTION: 2, MALFORMED_DATA: 1}
I would use dict comprehension
d={getErrorType(a):a for a in arr}
EDIT:
ok to get the counts like OP said, i would do something like this
d={x: [getErrorType(a) for a in arr].count(x) for x in set([getErrorType(a) for a in arr])}
Though this may make it too hard to read to be pythonic/////
data = {}
for a in arr:
error_type = getErrorType(a)
if error_type in data:
data[error_type] += 1
else:
data[error_type] = 1
Don't think there is an efficient method to keep count using dict comprehension.
I would probably just use an iterative approach using a normal dictionary or a defaultdict
from collections import defaultdict
d = defaultdict(int)
for num in arr: d[getErrorType(num)] += 1
Made a generic function just to simulate you could pass the whole arr into your function then you could use .count on the new list that has the results to form a dicitonary
def getErrorType(a):
return ['Ex' if i % 2 else 'Mal' for i in a ]
arr = [1, 2, 3]
lista = getErrorType(arr)
dicta = {i: lista.count(i) for i in lista}
(xenial)vash#localhost:~/python/stack_overflow$ python3.7 helping.py
{'Ex': 2, 'Mal': 1}
I do not agree with looping through += 1 every item to create a dictionary that doesn't seem efficient, I stand by this
Combined some solution above
d = defaultdict(int)
for num in set(arr):
d[getErrorType(num)] += arr.count(num)
I have my_dict with sets as values and I have x which is also a set.
I need to return list with set from my dict which contain all numbers in x. If set in my_dict does not contain all numbers in x I do not want to return it.
I want to use intersection (&) but it returns all the sets in my_dict.
my_dict = {1: {1,2,3,4,5},
2: {1,2,3,7,8},
3: {1,2,3,4}
}
x = {1,2,5}
new_list = []
for i in my_dict:
if my_dict[i] & x:
new_list.append(i)
print(new_list)
Output:
[1, 2, 3]
I need to receive [1] instead of [1, 2, 3]
When intersection becomes x that means all values in x are present in the set in dictionary.
for i in my_dict:
if (my_dict[i] & x)==x:
new_list.append(i)
print(new_list)
Edit: as suggested in the comments below you can also do
for i in my_dict:
if x.issubset(my_dict[i]):
new_list.append(i)
print(new_list)
I suggest you use the set.issuperset method, rather than using the & operator. Why combine several operators when a method exists to do exactly what you want?
new_list = []
for i in my_dict:
if my_dict[i].issuperset(x):
new_list.append(i)
Note that I'd normally write this with a list comprehension:
newlist = [key for key, value in my_dict.items() if value.issuperset(x)]
The inter section between my_dict values and x should be equal to x that means x should be a subset of my_dict value
my_dict = {1: {1,2,3,4,5},
2: {1,2,3,7,8},
3: {1,2,3,4}}
x = {1,2,5}
new_list = []
for i,j in my_dict.items():
if x.issubset(j):
new_list.append(i)
print(new_list)
This can also be solved using the issubset function. Here's an example:
for i in my_dict:
if x.issubset(my_dict[i]):
new_list.append(i)
Output: [1]
In this example, we're checking whether the value of every key value pair in the dictionary is a super-set of x (in other words x belongs to my_dict[i]), if that is the case then we just append the index to the desired list.
To check whether the entirety of a set is within another set, the nicest (in my opinon) way is to use the < and > operators, which are override to act as the equivalent of "is a superset of" in mathematics, and equivalent to the set.issuperset method. The advantage of this way is that the >= and <= operators are naturally available to check non-strict supersets.
Here's quite an idomatic way of doing it:
new_list = []
for key, value in my_dict.items():
if value >= x:
new_list.append(key)
The problem with your original code is it checks to see if there is any intersection between the two sets, i.e. they share even just one element, when you seem to want to check if all of x: set is in the set you're checking against.
I would also advise using a list compehension if you want to simplify the code, unless you have other steps you also need to do.
new_list = [key for key, value in my_dict.items() if value >= x]
I have a list populated from entries of a log; for sake of simplicity, something like
listlog = ["entry1:abcde", "entry2:abbds", "entry1:eorieo", "entry3:orieqor", "entry2:iroewiow"......]
This list can have an undefined number of entry, which may or may not be in sequence, since I run multiple operations in async fashion.
Then I have another list, which I use as reference to get only the list of entries; which may be like
list_template = ["entry1", "entry2", "entry3"]
I am trying to use the second list, to get sequences of entries, so I can isolate the single sequence, taking only the first instance found of each entry.
Since I am not dealing with numbers, I can't use set, so I did try with a loop inside a loop, comparing values in each list
This does not work, because it is possible that another entry may happen before what I am looking for (say, I want entry1, entry2, entry3, and the loop find entry1, but then find entry3, and since I compare every element of each list, it will be happy to find an element)
for item in listlog:
entry, value = item.split(":")
for reference_entry in list_template:
if entry == reference_entry:
print item
break
I have to, in a nutshell, find a sequence as in the template list, while these items are not necessarily in order. I am trying to parse the list once, otherwise I could do a very expensive multi-pass for each element of the template list, until I find the first occurrence and bail out. I thought that doing the loop in the loop is more efficient, since my reference list is always smaller than the log list, which is usually few elements.
How would you approach this problem, in the most efficient and pythonic way? All that I can think of, is multiple passes on the log list
you can use dict:
>>> listlog
['entry1:abcde', 'entry2:abbds', 'entry1:eorieo', 'entry3:orieqor', 'entry2:iroewiow']
>>> list_template
['entry1', 'entry2', 'entry3']
>>> for x in listlog:
... key, value = x.split(":")
... if key not in my_dict and key in list_template:
... my_dict[key] = value
...
>>> my_dict
{'entry2': 'abbds', 'entry3': 'orieqor', 'entry1': 'abcde'}
Disclaimer : This answer could use someone's insight on performance. Sure, list/dict comprehensions and zip are pythonic but the following may very well be a poor use of those tools.
You could use zip :
>>> data = ["a:12", "b:32", "c:54"]
>>> ref = ['c', 'b']
>>> matches = zip(ref, [val for key,val in [item.split(':') for item in data] if key in ref])
>>> for k, v in matches:
>>> print("{}:{}".format(k, v))
c:32
b:54
Here's another (worse? I'm not sure, performance-wise) way to get around this :
>>> data = ["a:12", "b:32", "c:54"]
>>> data_dict = {x:y for x,y in [item.split(':') for item in data]}
>>> ["{}:{}".format(key, val) for key,val in md.items() if key in ref]
['b:32', 'c:54']
Explanation :
Convert your initial list into a dict using a dict
For each pair of (key, val) found in the dict, join both in a string if the key is found in the 'ref' list
You can use a list comprehension something like this:
import re
listlog = ["entry1:abcde", "entry2:abbds", "entry1:eorieo", "entry3:orieqor", "entry2:iroewiow"]
print([item for item in listlog if re.search('entry', item)])
# ['entry1:abcde', 'entry2:abbds', 'entry1:eorieo', 'entry3:orieqor', 'entry2:iroewiow']
Than u can split 'em as u wish and create a dictonary if u want:
import re
listlog = ["entry1:abcde", "entry2:abbds", "entry1:eorieo", "entry3:orieqor", "entry2:iroewiow"]
mylist = [item for item in listlog if re.search('entry', item)]
def create_dict(string, dict_splitter=':'):
_dict = {}
temp = string.split(dict_splitter)
key = temp[0]
value = temp[1]
_dict[key] = value
return _dict
mydictionary = {}
for x in mylist:
x = str(x)
mydictionary.update(create_dict(x))
for k, v in mydictionary.items():
print(k, v)
# entry1 eorieo
# entry2 iroewiow
# entry3 orieqor
As you see this method need an update, cause we have changing the dictionary value. That's bad. Most better to update value for the same key. But it's much easier as u can think
I defined a dictionary like this (list is a list of integers):
my_dictionary = {'list_name' : list, 'another_list_name': another_list}
Now, I want to create a new list by iterating over this dictionary. In the end, I want it to look like this:
my_list = [list_name_list_item1, list_name_list_item2,
list_name_list_item3, another_list_name_another_list_item1]
And so on.
So my question is: How can I realize this?
I tried
for key in my_dictionary.keys():
k = my_dictionary[key]
for value in my_dictionary.values():
v = my_dictionary[value]
v = str(v)
my_list.append(k + '_' + v)
But instead of the desired output I receive a Type Error (unhashable type: 'list') in line 4 of this example.
You're trying to get a dictionary item by it's value whereas you already have your value.
Do it in one line using a list comprehension:
my_dictionary = {'list_name' : [1,4,5], 'another_list_name': [6,7,8]}
my_list = [k+"_"+str(v) for k,lv in my_dictionary.items() for v in lv]
print(my_list)
result:
['another_list_name_6', 'another_list_name_7', 'another_list_name_8', 'list_name_1', 'list_name_4', 'list_name_5']
Note that since the order in your dictionary is not guaranteed, the order of the list isn't either. You could fix the order by sorting the items according to keys:
my_list = [k+"_"+str(v) for k,lv in sorted(my_dictionary.items()) for v in lv]
Try this:
my_list = []
for key in my_dictionary:
for item in my_dictionary[key]:
my_list.append(str(key) + '_' + str(item))
Hope this helps.
Your immediate problem is that dict().values() is a generator yielding the values from the dictionary, not the keys, so when you attempt to do a lookup on line 4, it fails (in this case) as the values in the dictionary can't be used as keys. In another case, say {1:2, 3:4}, it would fail with a KeyError, and {1:2, 2:1} would not raise an error, but likely give confusing behaviour.
As for your actual question, lists do not attribute any names to data, like dictionaries do; they simply store the index.
def f()
a = 1
b = 2
c = 3
l = [a, b, c]
return l
Calling f() will return [1, 2, 3], with any concept of a, b, and c being lost entirely.
If you want to simply concatenate the lists in your dictionary, making a copy of the first, then calling .extend() on it will suffice:
my_list = my_dictionary['list_name'][:]
my_list.extend(my_dictionary['another_list_name'])
If you're looking to keep the order of the lists' items, while still referring to them by name, look into the OrderedDict class in collections.
You've written an outer loop over keys, then an inner loop over values, and tried to use each value as a key, which is where the program failed. Simply use the dictionary's items method to iterate over key,value pairs instead:
["{}_{}".format(k,v) for k,v in d.items()]
Oops, failed to parse the format desired; we were to produce each item in the inner list. Not to worry...
d={1:[1,2,3],2:[4,5,6]}
list(itertools.chain(*(
["{}_{}".format(k,i) for i in l]
for (k,l) in d.items() )))
This is a little more complex. We again take key,value pairs from the dictionary, then make an inner loop over the list that was the value and format those into strings. This produces inner sequences, so we flatten it using chain and *, and finally save the result as one list.
Edit: Turns out Python 3.4.3 gets quite confused when doing this nested as generator expressions; I had to turn the inner one into a list, or it would replace some combination of k and l before doing the formatting.
Edit again: As someone posted in a since deleted answer (which confuses me), I'm overcomplicating things. You can do the flattened nesting in a chained comprehension:
["{}_{}".format(k,v) for k,l in d.items() for v in l]
That method was also posted by Jean-François Fabre.
Use list comprehensions like this
d = {"test1":[1,2,3,],"test2":[4,5,6],"test3":[7,8,9]}
new_list = [str(item[0])+'_'+str(v) for item in d.items() for v in item[1]]
Output：
new_list:
['test1_1',
'test1_2',
'test1_3',
'test3_7',
'test3_8',
'test3_9',
'test2_4',
'test2_5',
'test2_6']
Let's initialize our data
In [1]: l0 = [1, 2, 3, 4]
In [2]: l1 = [10, 20, 30, 40]
In [3]: d = {'name0': l0, 'name1': l1}
Note that in my example, different from yours, the lists' content is not strings... aren't lists heterogeneous containers?
That said, you cannot simply join the keys and the list's items, you'd better cast these value to strings using the str(...) builtin.
Now it comes the solution to your problem... I use a list comprehension
with two loops, the outer loop comes first and it is on the items (i.e., key-value couples) in the dictionary, the inner loop comes second and it is on the items in the corresponding list.
In [4]: res = ['_'.join((str(k), str(i))) for k, l in d.items() for i in l]
In [5]: print(res)
['name0_1', 'name0_2', 'name0_3', 'name0_4', 'name1_10', 'name1_20', 'name1_30', 'name1_40']
In [6]:
In your case, using str(k)+'_'+str(i) would be fine as well, but the current idiom for joining strings with a fixed 'text' is the 'text'.join(...) method. Note that .join takes a SINGLE argument, an iterable, and hence in the list comprehension I used join((..., ...))
to collect the joinands in a single argument.