get keys correspond to a value in dictionary

Question

I have this dictionary in python

d={1: 'a', 2: 'b', 3: 'c'}

with d[1] I get

>>> d[1]
'a'

how I can get the key correspond to a value ?
example : with 'a' get 1

Answer 1

Key = next((x for x in d if d[x] == 'a'), None)

This gives the desired result if only one corresponding key exists. (In O(n) time?) Since it quits immediately once it finds a qualified key.

Keys = [x for x in d if d[x] == 'a']

gives all qualified keys (in a list). And it returns an empty list if there are no such keys. This method is called 'list comprehension'.

Avoid getting keys by values whenever possible. It's quite inefficient comparing to accessing values from keys in O(1).

This implies that you wanna create a inversely mapping dict where keys and values of old dict are switched as values and keys in the new dict -- if you are gonna use this method frequently thereafter.

Answer 2

In the case you are looking for getting only one value, there's probably a faster way to do that. If you are sure that the value exists in your dictionnary, then :

my_value = 'a'
return next(key for key, value in d.iteritems() if value == my_value)

If you are not sure about getting a result, you can surround that by a try/except catching the StopIteration exception.

Here, dictionnary's items are generated until the value match. If you have a very large dictionnary, this trick may be faster in some cases.

---

Using a large set of data :

d1={1: 1, 2: 2, 3: 8, 4: 3, 5: 6, 6: 9, 7: 17, 8: 4, 9: 20, 10: 7, 11: 15,
    12: 10, 13: 10, 14: 18, 15: 18, 16: 5, 17: 13, 18: 21, 19: 21, 20: 8,
    21: 8, 22: 16, 23: 16, 24: 11, 25: 24, 26: 11, 27: 112, 28: 19, 29: 19,
    30: 19, 3077: 36, 32: 6, 33: 27, 34: 14, 35: 14, 36: 22, 4102: 39, 38: 22,
    39: 35, 40: 9, 41: 110, 42: 9, 43: 30, 44: 17, 45: 17, 46: 17, 47: 105, 48: 12,
    49: 25, 50: 25, 51: 25, 52: 12, 53: 12, 54: 113, 1079: 50, 56: 20, 57: 33,
    58: 20, 59: 33, 60: 20, 61: 20, 62: 108, 63: 108, 64: 7, 65: 28, 66: 28, 67: 28,
    68: 15, 69: 15, 70: 15, 71: 103, 72: 23, 73: 116, 74: 23, 75: 15, 76: 23, 77: 23, 
    78: 36, 79: 36, 80: 10, 81: 23, 82: 111, 83: 111, 84: 10, 85: 10, 86: 31, 87: 31, 
    88: 18, 89: 31, 90: 18, 91: 93, 92: 18, 93: 18, 94: 106, 95: 106, 96: 13, 9232: 35, 
    98: 26, 99: 26, 100: 26, 101: 26, 103: 88, 104: 13, 106: 13, 107: 101, 1132: 63, 
    2158: 51, 112: 21, 113: 13, 116: 21, 118: 34, 119: 34, 7288: 45, 121: 96, 122: 21, 
    124: 109, 125: 109, 128: 8, 1154: 32, 131: 29, 134: 29, 136: 16, 137: 91, 140: 16, 
    142: 104, 143: 104, 146: 117, 148: 24, 149: 24, 152: 24, 154: 24, 155: 86, 160: 11, 
    161: 99, 1186: 76, 3238: 49, 167: 68, 170: 11, 172: 32, 175: 81, 178: 32, 179: 32, 
    182: 94, 184: 19, 31: 107, 188: 107, 190: 107, 196: 27, 197: 27, 202: 27, 206: 89, 
    208: 14, 214: 102, 215: 102, 220: 115, 37: 22, 224: 22, 226: 14, 232: 22, 233: 84, 
    238: 35, 242: 97, 244: 22, 250: 110, 251: 66, 1276: 58, 256: 9, 2308: 33, 262: 30, 
    263: 79, 268: 30, 269: 30, 274: 92, 1300: 27, 280: 17, 283: 61, 286: 105, 292: 118, 
    296: 25, 298: 25, 304: 25, 310: 87, 1336: 71, 319: 56, 322: 100, 323: 100, 325: 25, 
    55: 113, 334: 69, 340: 12, 1367: 40, 350: 82, 358: 33, 364: 95, 376: 108, 
    377: 64, 2429: 46, 394: 28, 395: 77, 404: 28, 412: 90, 1438: 53, 425: 59, 430: 103, 
    1456: 97, 433: 28, 445: 72, 448: 23, 466: 85, 479: 54, 484: 98, 485: 98, 488: 23, 
    6154: 37, 502: 67, 4616: 34, 526: 80, 538: 31, 566: 62, 3644: 44, 577: 31, 97: 119, 
    592: 26, 593: 75, 1619: 48, 638: 57, 646: 101, 650: 26, 110: 114, 668: 70, 2734: 41, 
    700: 83, 1732: 30, 719: 52, 728: 96, 754: 65, 1780: 74, 4858: 47, 130: 29, 790: 78, 
    1822: 43, 2051: 38, 808: 29, 850: 60, 866: 29, 890: 73, 911: 42, 958: 55, 970: 99, 
    976: 24, 166: 112}

I compared the following functions in order to check which one was the fastest.

def f1():  
    return next(key for key, value in d1.iteritems() if value == 55)

def f2a():
    return [item[0] for item in d1.items() if item[1] == 55][0]

def f2b():
    return [item[0] for item in d1.iteritems() if item[1] == 55][0]

def f3():
    return dict((v, k) for k, v in d1.iteritems())[55]

tl=["f1", "f2a", "f2b", "f3"]
from timeit import Timer
cmpthese(1000, tl)

And here are the results :

f3   21182/s  --   -39%   -40%   -56%
f2a  34450/s    63%  --    -3%   -29%
f2b  35368/s    67%     3%  --   -27%
f1   48367/s   128%    40%    37%  --

f1 is faster that any other one.

Answer 3

key = 'a'
return [item[0] for item in self.items() if item[1] == key]

This will find all keys with value 'a' and return a list of them.

Answer 4

You want to invert the mapping:

As shown in this question

dict((v,k) for k, v in map.iteritems())

Answer 5

You could create a new dictionary from the keys and values in the initial one:

>>> d2 = dict((v, k) for k, v in d.iteritems())
>>> d2
{'a': 1, 'c': 3, 'b': 2}
>>> d2['a']
1