Related
Given:
a_dict = {3: [3, 7], 7: [6, 2], 2: [2, 1], 8: [-3, 9]}
I'm stuck on a problem where I have to return an updated dictionary that contains the sum of the elements inside the list values alongside its key.
For example, the first key pair:
3: [3, 7]
is to be turned into this:
3: 10
Returned updated dictionary should look like this:
{3: 10, 7: 8, 2: 3, 8: 6}
I'm not sure if I should use the values method to isolate the values to get the sum of numbers from the list values.
for key, val in a_dict.items():
a_dict[key] = sum(val)
Use a dict comprehension:
{k: sum(v) for k, v in a_dict.items()}
# {3: 10, 7: 8, 2: 3, 8: 6}
in python, if I want to find the max value of d, but the key only include 1,2,3 other than all the keys in the d. so how to do, thank you.
d = {1: 5, 2: 0, 3: 4, 4: 0, 5: 1}
Just get the keys and values for the keys 1, 2 and 3 in a list of tuples, sort the list and get the first tuple element [0] key [0].
d = {1: 5, 2: 0, 3: 4, 4: 0, 5: 1}
key_max_val = sorted([(k,v) for k,v in d.items() if k in [1,2,3]])[0][0]
print(key_max_val) # Outputs 1
You can use operator:
It will return you the key with maximum value:
In [873]: import operator
In [874]: d = {1: 5, 2: 0, 3: 4, 4: 0, 5: 1}
In [875]: max(d.iteritems(), key=operator.itemgetter(1))[0]
Out[875]: 1
I think this below should work (base on
#Mayank Porwal idea, sorry coz I can not reply):
d = {1: 5, 2: 0, 3: 4, 4: 0, 5: 1}
max(v for k,v in d.items())
Use a generator and the max builtin function:
Max value
max(v for k,v in d.items() if k in [1,2,3])
Max key
max(k for k,v in d.items() if k in [1,2,3])
If you have a dictionary of integers:
d = {
1:[0],
2:[1],
3:[0,1,2,3,4],
4:[0],
5:[1],
6:[0,1,2,3,4],
11:[0],
22:[1],
33:[0,1,2,3,4],
44:[0],
55:[1],
66:[0,1,2,3,4]
}
You want to:
Validate that the keys are between 0 and 25.
Delete any keys that are outside of the range as they are not valid and will ruin the data set.
Dictionary keys are not naturally sorted.
Given, how would validate that your keys are in the required range?
My try:
for x,y in d.items():
if x<0 or x>25:
del d[x]
When ran I get the error:
RuntimeError: dictionary changed size during iteration
How would I compensate for this?
In your example, you are mutating the d while looping through it. This is bad.
The easiest way to do this if you don't need to change the original d is to use a dictionary comprehension:
d = {k: v for k, v in d.items() if 0 <= k <= 25}
If you want to delete keys while iterating, you need to iterate over a copy instead and pop keys that don't hold to your condition:
d = {1:[0], 2:[1], 3:[0,1,2,3,4], 4:[0], 5:[1], 6:[0,1,2,3,4], 11:[0], 22:[1], 33:[0,1,2,3,4], 44:[0], 55:[1], 66:[0,1,2,3,4]}
for k in d.copy(): # or list(d)
if not 0 <= k <= 25:
d.pop(k) # or del d[k]
Which Outputs:
{1: [0], 2: [1], 3: [0, 1, 2, 3, 4], 4: [0], 5: [1], 6: [0, 1, 2, 3, 4], 11: [0], 22: [1]}
As others have shown, reconstructing a new dictionary is always an easy way around this.
You can use a basic dict comprehension here:
{k: d[k] for k in d if 0 <= k <= 25}
Or even a functional approach with filter():
dict(filter(lambda x: 0 <= x[0] <= 25, d.items()))
You can use a dictionary comprehension:
d = { 1:[0], 2:[1], 3:[0,1,2,3,4], 4:[0], 5:[1], 6:[0,1,2,3,4], 11:[0], 22:[1], 33:[0,1,2,3,4], 44:[0], 55:[1], 66:[0,1,2,3,4] }
new_d = {a:b for a, b in d.items() if a <= 25 and a >= 0}
Output:
{1: [0], 2: [1], 3: [0, 1, 2, 3, 4], 4: [0], 5: [1], 6: [0, 1, 2, 3, 4], 11: [0], 22: [1]}
I have a list: [1,2,3,4,5,6,0,9,2] that I want to loop through.
I want to create another list of dictionaries for every 3 elements until I reach a 0 element (i.e. the first dictionary will be {'gold':1,'silver':2,'bronze':3} and similarly for 4,5,6). The loop will terminate since it reaches 0.
So the resulting list would be: [{'gold':1,'silver':2,'bronze':3},{'gold':4,'silver':5,'bronze':6}]
Being new to Python and coming from a PHP background, how would I solve this leveraging functions like zip or map or would I have to resort to plain while loops?
from itertools import cycle
placings = cycle(['gold', 'silver', 'bronze'])
numbers = [1,2,3,4,5,6,0,9,2]
def zipper(numbers, placings):
d = {}
for n, c in zip(numbers, placings):
if n == 0:
break
d[n] = c
if n % 3 == 0:
yield d
d = {}
print list(zipper(numbers, placings))
from itertools import takewhile, islice
def groups(l, keys, i, n):
# lambda x: x != 0 -> take elements until we encounter any 0
tk = takewhile(lambda x: x != i, l)
while True:
d = dict(zip(keys, islice(tk, n)))
if len(d) == n: # make sure we have n keys in the dict
yield d
else:
break
keys = ['gold', 'silver', 'bronze']
print(list(groups(l,keys,0,3)))
[{'bronze': 3, 'gold': 1, 'silver': 2}, {'bronze': 6, 'gold': 4, 'silver': 5}]
takewhile
Make an iterator that returns elements from the iterable as long as the predicate is true.
So we only take elements up to the first i we encounter, d = islice(tk,n) takes a n length slice from our takewhile object using islice, we then create a dict from the passed in keys, Creating the dict consumes n elements at a time so we move to the next three elements each time we call it. If the length of the takewhile object is not a multiple of of n then we will have a dict at the end that is less than the size of n so if len(d) == n: will catch that and break the loop so the odd the length dict or an empty dict will terminate the while.
If you allow dicts with less than three keys:
from itertools import takewhile, islice,chain
def groups(l, keys, i, n):
tk = takewhile(lambda x: x != i, l)
return (dict(zip(keys, chain(islice(tk, n - 1),(i,)))) for i in tk)
Output:
In [5]: l = [1, 2, 3, 4, 5, 6, 7, 11, 6, 9, 0, 2]
In [6]: list(groups(l,['gold', 'silver', 'bronze'],0,3))
Out[6]:
[{'bronze': 1, 'gold': 2, 'silver': 3},
{'bronze': 4, 'gold': 5, 'silver': 6},
{'bronze': 7, 'gold': 11, 'silver': 6},
{'gold': 9}]
islice(tk, n - 1) takes a slice from our takewhile object len n - 1, we chain that and i so we get n elements each time.
Or use filter to filter out the dict with less than 3 keys:
def groups(l, keys, i, n):
tk = takewhile(lambda x: x != i, l)
return filter(lambda x: len(x) == n, (dict(zip(keys, chain(islice(tk, n - 1),(i,))))
for i in tk))
Which removes the shorter dict:
In [8]: list(groups(l,['gold', 'silver', 'bronze'],0,3))
Out[8]:
[{'bronze': 1, 'gold': 2, 'silver': 3},
{'bronze': 4, 'gold': 5, 'silver': 6},
{'bronze': 7, 'gold': 11, 'silver': 6}]
This does the same as the previous we just filter any potential odd length dict from the end. For python2 use itertools.ifilter instead of filter.
Easy! With a simple function, a data structure and simple iteration:
Example:
from collections import defaultdict
def f(xs, labels=("gold", "silver", "bronze")):
d = defaultdict(int)
n = 0
for i, x in enumerate(xs):
if x == 0: # break if we see a 0
return d
d[labels[n]] += i # update our counts
# cycle through gold/silver/bronze (labels)
n += 1
if n % 3 == 0:
n = 0
return d
xs = [1, 2, 3, 4, 5, 6, 0, 9, 2]
d = f(xs)
print d
Output:
$ python foo.py
defaultdict(<type 'int'>, {'bronze': 7, 'silver': 5, 'gold': 3})
First of all you need to break the larger list into smaller lists of same size as of list medals, to make the zip() function work correctly, Also the position of 0 in the list should be at a position divisible by the length of the medals list, Otherwise the last list among the smaller lists would have lesser number of elements.
listOfDictionary = []
lst = [1,2,3,4,5,6,0,9,2]
ind = lst.index(0) #Tells where to break the iteration loop.
medals = ["gold", "silver", "bronze"]
if ind%3==0:
newlst = [lst[i: i+3]for i in xrange(0,ind,3)]
#newlst = [[1, 2, 3], [4, 5, 6]]
for lsst in newlst:
listOfDictionary.append(dict(zip(medals, lsst)))
print listOfDictionary
>>> [{'bronze': 3, 'silver': 2, 'gold': 1}, {'bronze': 6, 'silver': 5, 'gold': 4}]
Using list comprehension in case of list and then using cycle
In [61]: mykeys
Out[61]: [1, 2, 3, 4, 5, 6, 0, 9, 2]
In [63]: mykeys1=[i for i in mykeys[:-1 if 0 not in mykeys else mykeys.index(0)]]
In [64]: myvalues
Out[64]: ['gold', 'silver', 'bronze']
In [66]: myvalues1=cycle(myvalues)
In [67]: mydict=dict(zip(mykeys1,myvalues1))
In [68]: mydict
Out[68]: {1: 'gold', 2: 'silver', 3: 'bronze', 4: 'gold', 5: 'silver', 6: 'bronze'}
In [69]: [dict(mydict.items()[i:i+3]) for i in range(0,len(mykeys1),3)]
Out[69]: [{1: 'gold', 2: 'silver', 3: 'bronze'}, {4: 'gold', 5: 'silver', 6: 'bronze'}]
I want to change back and forth between a dictionary of (equal-length) lists:
DL = {'a': [0, 1], 'b': [2, 3]}
and a list of dictionaries:
LD = [{'a': 0, 'b': 2}, {'a': 1, 'b': 3}]
For those of you that enjoy clever/hacky one-liners.
Here is DL to LD:
v = [dict(zip(DL,t)) for t in zip(*DL.values())]
print(v)
and LD to DL:
v = {k: [dic[k] for dic in LD] for k in LD[0]}
print(v)
LD to DL is a little hackier since you are assuming that the keys are the same in each dict. Also, please note that I do not condone the use of such code in any kind of real system.
If you're allowed to use outside packages, Pandas works great for this:
import pandas as pd
pd.DataFrame(DL).to_dict(orient="records")
Which outputs:
[{'a': 0, 'b': 2}, {'a': 1, 'b': 3}]
You can also use orient="list" to get back the original structure
{'a': [0, 1], 'b': [2, 3]}
Perhaps consider using numpy:
import numpy as np
arr = np.array([(0, 2), (1, 3)], dtype=[('a', int), ('b', int)])
print(arr)
# [(0, 2) (1, 3)]
Here we access columns indexed by names, e.g. 'a', or 'b' (sort of like DL):
print(arr['a'])
# [0 1]
Here we access rows by integer index (sort of like LD):
print(arr[0])
# (0, 2)
Each value in the row can be accessed by column name (sort of like LD):
print(arr[0]['b'])
# 2
To go from the list of dictionaries, it is straightforward:
You can use this form:
DL={'a':[0,1],'b':[2,3], 'c':[4,5]}
LD=[{'a':0,'b':2, 'c':4},{'a':1,'b':3, 'c':5}]
nd={}
for d in LD:
for k,v in d.items():
try:
nd[k].append(v)
except KeyError:
nd[k]=[v]
print nd
#{'a': [0, 1], 'c': [4, 5], 'b': [2, 3]}
Or use defaultdict:
nd=cl.defaultdict(list)
for d in LD:
for key,val in d.items():
nd[key].append(val)
print dict(nd.items())
#{'a': [0, 1], 'c': [4, 5], 'b': [2, 3]}
Going the other way is problematic. You need to have some information of the insertion order into the list from keys from the dictionary. Recall that the order of keys in a dict is not necessarily the same as the original insertion order.
For giggles, assume the insertion order is based on sorted keys. You can then do it this way:
nl=[]
nl_index=[]
for k in sorted(DL.keys()):
nl.append({k:[]})
nl_index.append(k)
for key,l in DL.items():
for item in l:
nl[nl_index.index(key)][key].append(item)
print nl
#[{'a': [0, 1]}, {'b': [2, 3]}, {'c': [4, 5]}]
If your question was based on curiosity, there is your answer. If you have a real-world problem, let me suggest you rethink your data structures. Neither of these seems to be a very scalable solution.
Here are the one-line solutions (spread out over multiple lines for readability) that I came up with:
if dl is your original dict of lists:
dl = {"a":[0, 1],"b":[2, 3]}
Then here's how to convert it to a list of dicts:
ld = [{key:value[index] for key,value in dl.items()}
for index in range(max(map(len,dl.values())))]
Which, if you assume that all your lists are the same length, you can simplify and gain a performance increase by going to:
ld = [{key:value[index] for key, value in dl.items()}
for index in range(len(dl.values()[0]))]
Here's how to convert that back into a dict of lists:
dl2 = {key:[item[key] for item in ld]
for key in list(functools.reduce(
lambda x, y: x.union(y),
(set(dicts.keys()) for dicts in ld)
))
}
If you're using Python 2 instead of Python 3, you can just use reduce instead of functools.reduce there.
You can simplify this if you assume that all the dicts in your list will have the same keys:
dl2 = {key:[item[key] for item in ld] for key in ld[0].keys() }
cytoolz.dicttoolz.merge_with
Docs
from cytoolz.dicttoolz import merge_with
merge_with(list, *LD)
{'a': [0, 1], 'b': [2, 3]}
Non-cython version
Docs
from toolz.dicttoolz import merge_with
merge_with(list, *LD)
{'a': [0, 1], 'b': [2, 3]}
The python module of pandas can give you an easy-understanding solution. As a complement to #chiang's answer, the solutions of both D-to-L and L-to-D are as follows:
import pandas as pd
DL = {'a': [0, 1], 'b': [2, 3]}
out1 = pd.DataFrame(DL).to_dict('records')
Output:
[{'a': 0, 'b': 2}, {'a': 1, 'b': 3}]
In the other direction:
LD = [{'a': 0, 'b': 2}, {'a': 1, 'b': 3}]
out2 = pd.DataFrame(LD).to_dict('list')
Output:
{'a': [0, 1], 'b': [2, 3]}
Cleanest way I can think of a summer friday. As a bonus, it supports lists of different lengths (but in this case, DLtoLD(LDtoDL(l)) is no more identity).
From list to dict
Actually less clean than #dwerk's defaultdict version.
def LDtoDL (l) :
result = {}
for d in l :
for k, v in d.items() :
result[k] = result.get(k,[]) + [v] #inefficient
return result
From dict to list
def DLtoLD (d) :
if not d :
return []
#reserve as much *distinct* dicts as the longest sequence
result = [{} for i in range(max (map (len, d.values())))]
#fill each dict, one key at a time
for k, seq in d.items() :
for oneDict, oneValue in zip(result, seq) :
oneDict[k] = oneValue
return result
I needed such a method which works for lists of different lengths (so this is a generalization of the original question). Since I did not find any code here that the way that I expected, here's my code which works for me:
def dict_of_lists_to_list_of_dicts(dict_of_lists: Dict[S, List[T]]) -> List[Dict[S, T]]:
keys = list(dict_of_lists.keys())
list_of_values = [dict_of_lists[key] for key in keys]
product = list(itertools.product(*list_of_values))
return [dict(zip(keys, product_elem)) for product_elem in product]
Examples:
>>> dict_of_lists_to_list_of_dicts({1: [3], 2: [4, 5]})
[{1: 3, 2: 4}, {1: 3, 2: 5}]
>>> dict_of_lists_to_list_of_dicts({1: [3, 4], 2: [5]})
[{1: 3, 2: 5}, {1: 4, 2: 5}]
>>> dict_of_lists_to_list_of_dicts({1: [3, 4], 2: [5, 6]})
[{1: 3, 2: 5}, {1: 3, 2: 6}, {1: 4, 2: 5}, {1: 4, 2: 6}]
>>> dict_of_lists_to_list_of_dicts({1: [3, 4], 2: [5, 6], 7: [8, 9, 10]})
[{1: 3, 2: 5, 7: 8},
{1: 3, 2: 5, 7: 9},
{1: 3, 2: 5, 7: 10},
{1: 3, 2: 6, 7: 8},
{1: 3, 2: 6, 7: 9},
{1: 3, 2: 6, 7: 10},
{1: 4, 2: 5, 7: 8},
{1: 4, 2: 5, 7: 9},
{1: 4, 2: 5, 7: 10},
{1: 4, 2: 6, 7: 8},
{1: 4, 2: 6, 7: 9},
{1: 4, 2: 6, 7: 10}]
Here my small script :
a = {'a': [0, 1], 'b': [2, 3]}
elem = {}
result = []
for i in a['a']: # (1)
for key, value in a.items():
elem[key] = value[i]
result.append(elem)
elem = {}
print result
I'm not sure that is the beautiful way.
(1) You suppose that you have the same length for the lists
Here is a solution without any libraries used:
def dl_to_ld(initial):
finalList = []
neededLen = 0
for key in initial:
if(len(initial[key]) > neededLen):
neededLen = len(initial[key])
for i in range(neededLen):
finalList.append({})
for i in range(len(finalList)):
for key in initial:
try:
finalList[i][key] = initial[key][i]
except:
pass
return finalList
You can call it as follows:
dl = {'a':[0,1],'b':[2,3]}
print(dl_to_ld(dl))
#[{'a': 0, 'b': 2}, {'a': 1, 'b': 3}]
If you don't mind a generator, you can use something like
def f(dl):
l = list((k,v.__iter__()) for k,v in dl.items())
while True:
d = dict((k,i.next()) for k,i in l)
if not d:
break
yield d
It's not as "clean" as it could be for Technical Reasons: My original implementation did yield dict(...), but this ends up being the empty dictionary because (in Python 2.5) a for b in c does not distinguish between a StopIteration exception when iterating over c and a StopIteration exception when evaluating a.
On the other hand, I can't work out what you're actually trying to do; it might be more sensible to design a data structure that meets your requirements instead of trying to shoehorn it in to the existing data structures. (For example, a list of dicts is a poor way to represent the result of a database query.)
List of dicts ⟶ dict of lists
from collections import defaultdict
from typing import TypeVar
K = TypeVar("K")
V = TypeVar("V")
def ld_to_dl(ld: list[dict[K, V]]) -> dict[K, list[V]]:
dl = defaultdict(list)
for d in ld:
for k, v in d.items():
dl[k].append(v)
return dl
defaultdict creates an empty list if one does not exist upon key access.
Dict of lists ⟶ list of dicts
Collecting into "jagged" dictionaries
from typing import TypeVar
K = TypeVar("K")
V = TypeVar("V")
def dl_to_ld(dl: dict[K, list[V]]) -> list[dict[K, V]]:
ld = []
for k, vs in dl.items():
ld += [{} for _ in range(len(vs) - len(ld))]
for i, v in enumerate(vs):
ld[i][k] = v
return ld
This generates a list of dictionaries ld that may be missing items if the lengths of the lists in dl are unequal. It loops over all key-values in dl, and creates empty dictionaries if ld does not have enough.
Collecting into "complete" dictionaries only
(Usually intended only for equal-length lists.)
from typing import TypeVar
K = TypeVar("K")
V = TypeVar("V")
def dl_to_ld(dl: dict[K, list[V]]) -> list[dict[K, V]]:
ld = [dict(zip(dl.keys(), v)) for v in zip(*dl.values())]
return ld
This generates a list of dictionaries ld that have the length of the smallest list in dl.
DL={'a':[0,1,2,3],'b':[2,3,4,5]}
LD=[{'a':0,'b':2},{'a':1,'b':3}]
Empty_list = []
Empty_dict = {}
# to find length of list in values of dictionry
len_list = 0
for i in DL.values():
if len_list < len(i):
len_list = len(i)
for k in range(len_list):
for i,j in DL.items():
Empty_dict[i] = j[k]
Empty_list.append(Empty_dict)
Empty_dict = {}
LD = Empty_list