I use Python 3
I want to check if all of my tested values in the nested dictionary are non 0.
So here is the simplified example dict:
d = {'a': {'1990': 10, '1991': 0, '1992': 30},
'b': {'1990': 15, '1991': 40, '1992': 0}}
and I want to test if for both dicts 'a' and 'b' the values of the keys '1990' and '1991' are not zero
for i in d:
for k in range(2):
year = 1990
year = year + k
if all((d[i][str(year)]) != 0):
print(d[i])
so it should only return b, because a['1991']=0
but this is the first time I work with the all() function and I get the error core: TypeError: 'bool' object is not iterable
the error is in the if all() line
thank you very much!
This can done a bit more generally with a list comprehension where you iterate over the items in dict d. A simple comprehension to iterate over the keys and values in our dictionary looks like this:
>>> [k for k, v in d.items()]
['a', 'b']
In the above k will contain the keys and v the values. The comprehension also has an if clause. With that you can filter out the items you don't want. So we define years = ('1990', '1991'). Now we can do another comprehension to test our year values.
To iterate over only 'a', we could do this:
>>> [d['a'][y] for y in years]
[10, 0]
>>> all([d['a'][y] for y in years])
False
Gluing the whole thing together:
>>> d={'a' :{ '1990': 10, '1991':0, '1992':30},'b':{ '1990':15, '1991':40, '1992':0}}
>>> years = ('1990', '1991')
>>> [k for k, v in d.items() if all([v[y] for y in years])]
['b']
See the python docs for more information on list comprehensions.
Related
I have the following dictionary:
d = {}
d[1] = 'a'
d[2] = 'b'
d[3] = 'c'
d[4] = 'd'
I'd like to perform a reverse dictionary lookup for each character in a string:
input_string = "bad"
I get different results when I do this in a list comprehension as opposed to a nested for loop, and I don't understand why. As I understand, the list comprehension and the nested for loop should yield identical results. The list comprehension yields a list whose results are not in the order I would expect. My desired result here is that which is provided by the nested for loop, however I prefer to use the list comprehension to accomplish that. Perhaps this has something to do with python dictionary order of which I am unaware?
result1 = [key for key, value in d.items() for i in input_string if i == value]
print(result1)
> [1, 2, 4]
result2 = list()
for i in input_string:
for key, value in d.items():
if i == value:
result2.append(key)
print(result2)
> [2, 1, 4]
In order to mimic the traditional loop, the outer loop should be over input_string and the inner loop should be over d in the list comprehension:
out = [k for i in input_string for k,v in d.items() if i==v]
Output:
[2, 1, 4]
I have a 2d list with arbitrary strings like this:
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
I want to create a dictionary out of this:
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
How do I do this? This answer answers for 1D list for non-repeated values, but, I have a 2d list and values can repeat. Is there a generic way of doing this?
Maybe you could use two for-loops:
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
d = {}
overall_idx = 0
for sub_lst in lst:
for word in sub_lst:
if word not in d:
d[word] = overall_idx
# Increment overall_idx below if you want to only increment if word is not previously seen
# overall_idx += 1
overall_idx += 1
print(d)
Output:
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
You could first convert the list of lists to a list using a 'double' list comprehension.
Next, get rid of all the duplicates using a dictionary comprehension, we could use set for that but would lose the order.
Finally use another dictionary comprehension to get the desired result.
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
# flatten list of lists to a list
flat_list = [item for sublist in lst for item in sublist]
# remove duplicates
ordered_set = {x:0 for x in flat_list}.keys()
# create required output
the_dictionary = {v:i for i, v in enumerate(ordered_set)}
print(the_dictionary)
""" OUTPUT
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
"""
also, with collections and itertools:
import itertools
from collections import OrderedDict
lstdict={}
lst = [['a', 'xyz' , 'tps'], ['rtr' , 'xyz']]
lstkeys = list(OrderedDict(zip(itertools.chain(*lst), itertools.repeat(None))))
lstdict = {lstkeys[i]: i for i in range(0, len(lstkeys))}
lstdict
output:
{'a': 0, 'xyz': 1, 'tps': 2, 'rtr': 3}
d = {'A': ['A11117',
'33465'
'17160144',
'A11-33465',
'3040',
'A11-33465 W1',
'nor'], 'B': ['maD', 'vern', 'first', 'A2lRights']}
I have a dictionary d and I would like to sort the values based on length of characters. For instance, for key A the value A11-33465 W1 would be first because it contains 12 characters followed by 'A11-33465' because it contains 9 characters etc. I would like this output:
d = {'A': ['A11-33465 W1',
' A11-33465',
'17160144',
'A11117',
'33465',
'3040',
'nor'],
'B': ['A2lRights',
'first',
'vern',
'maD']}
(I understand that dictionaries are not able to be sorted but I have examples below that didn't work for me but the answer contains a dictionary that was sorted)
I have tried the following
python sorting dictionary by length of values
print(' '.join(sorted(d, key=lambda k: len(d[k]), reverse=True)))
Sort a dictionary by length of the value
sorted_items = sorted(d.items(), key = lambda item : len(item[1]))
newd = dict(sorted_items[-2:])
How do I sort a dictionary by value?
import operator
sorted_x = sorted(d.items(), key=operator.itemgetter(1))
But they both do not give me what I am looking for.
How do I get my desired output?
You are not sorting the dict, you are sorting the lists inside it. The simplest will be a loop that sorts the lists in-place:
for k, lst in d.items():
lst.sort(key=len, reverse=True)
This will turn d into:
{'A': ['3346517160144', 'A11-33465 W1', 'A11-33465', 'A11117', '3040', 'nor'],
'B': ['A2lRights', 'first', 'vern', 'maD']}
If you want to keep the original data intact, use a comprehension like:
sorted_d = {k: sorted(lst, key=len, reverse=True) for k, lst in d.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 can get this loop to work properly:
for x in range(0,len(l)):
for k in d:
if l[x] in d[k]:
l[x] = k
This looks through a list and checks if the value is in any of the dictionary items and then calculates it equal to the dictionary key it is found within (the dictionary contains lists.)
However, I want to convert to a list comprehension or other single line statement for use in a pandas dataframe - to populate a field based on whether or not another field's value is in the labeled dictionary keys and assign it the dictionary key value.
Here is my best attempt, but it does not work:
l = [ k for x in range(0,len(l)) if l[x] in d[k] for k in d ]
Thanks
Assuming I understand what you're after (example data that can be copied and pasted is always appreciated), I'd do something like this:
>>> l = ["a", "b", "c", "d"]
>>> d = {1: ["a"], 3: ["d", "c"]}
>>> l2 = [next((k for k,v in d.items() if lx in v), lx) for lx in l]
>>> l2
[1, 'b', 3, 3]
Don't forget to think about what behaviour you want if an entry in l is found in multiple lists in d, of course, although that may not be an issue with your data.
You can't do it with a list comprehension, because you have an assignment:
l[x] = k
which is an statement, and a list comprehension can't have them.