我有一个字典,其中键是字符串,值是整数。
stats = {'a': 1, 'b': 3000, 'c': 0}
如何获得具有最大值的键?在这种情况下,它是'b'。
有没有比使用带有反向键值元组的中间列表更好的方法?
inverse = [(value, key) for key, value in stats.items()]
print(max(inverse)[1])
我有一个字典,其中键是字符串,值是整数。
stats = {'a': 1, 'b': 3000, 'c': 0}
如何获得具有最大值的键?在这种情况下,它是'b'。
有没有比使用带有反向键值元组的中间列表更好的方法?
inverse = [(value, key) for key, value in stats.items()]
print(max(inverse)[1])
当前回答
max(stats, key=stats.get, default=None)
如果stats可以是一个空字典,则只使用max(stats, key=stats.get)将引发ValueError。
这个答案是安全的,只要在字典中不是一个可能的键。
其他回答
如果有一个以上的元素,我的值是最大的。我会列出所有值为最大值的键。
>>> stats = {'a':1000, 'b':3000, 'c': 100, 'd':3000}
>>> [key for m in [max(stats.values())] for key,val in stats.iteritems() if val == m]
['b', 'd']
这将给你'b'和任何其他最大键。
注意:对于python 3使用stats.items()而不是stats.iteritems()
如何:
max(zip(stats.keys(), stats.values()), key=lambda t : t[1])[0]
我已经测试了许多变体,这是返回最大值的dict键的最快方法:
def keywithmaxval(d):
""" a) create a list of the dict's keys and values;
b) return the key with the max value"""
v = list(d.values())
k = list(d.keys())
return k[v.index(max(v))]
为了给你一个概念,这里有一些候选方法:
def f1():
v = list(d1.values())
k = list(d1.keys())
return k[v.index(max(v))]
def f2():
d3 = {v: k for k,v in d1.items()}
return d3[max(d3)]
def f3():
return list(filter(lambda t: t[1] == max(d1.values()), d1.items()))[0][0]
def f3b():
# same as f3 but remove the call to max from the lambda
m = max(d1.values())
return list(filter(lambda t: t[1] == m, d1.items()))[0][0]
def f4():
return [k for k, v in d1.items() if v == max(d1.values())][0]
def f4b():
# same as f4 but remove the max from the comprehension
m = max(d1.values())
return [k for k,v in d1.items() if v == m][0]
def f5():
return max(d1.items(), key=operator.itemgetter(1))[0]
def f6():
return max(d1, key=d1.get)
def f7():
""" a) create a list of the dict's keys and values;
b) return the key with the max value"""
v = list(d1.values())
return list(d1.keys())[v.index(max(v))]
def f8():
return max(d1, key=lambda k: d1[k])
tl = [f1, f2, f3b, f4b, f5, f6, f7, f8, f4, f3]
cmpthese.cmpthese(tl, c=100)
测试词典:
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}
以及Python 3.2下的测试结果:
rate/sec f4 f3 f3b f8 f5 f2 f4b f6 f7 f1
f4 454 -- -2.5% -96.9% -97.5% -98.6% -98.6% -98.7% -98.7% -98.9% -99.0%
f3 466 2.6% -- -96.8% -97.4% -98.6% -98.6% -98.6% -98.7% -98.9% -99.0%
f3b 14,715 3138.9% 3057.4% -- -18.6% -55.5% -56.0% -56.4% -58.3% -63.8% -68.4%
f8 18,070 3877.3% 3777.3% 22.8% -- -45.4% -45.9% -46.5% -48.8% -55.5% -61.2%
f5 33,091 7183.7% 7000.5% 124.9% 83.1% -- -1.0% -2.0% -6.3% -18.6% -29.0%
f2 33,423 7256.8% 7071.8% 127.1% 85.0% 1.0% -- -1.0% -5.3% -17.7% -28.3%
f4b 33,762 7331.4% 7144.6% 129.4% 86.8% 2.0% 1.0% -- -4.4% -16.9% -27.5%
f6 35,300 7669.8% 7474.4% 139.9% 95.4% 6.7% 5.6% 4.6% -- -13.1% -24.2%
f7 40,631 8843.2% 8618.3% 176.1% 124.9% 22.8% 21.6% 20.3% 15.1% -- -12.8%
f1 46,598 10156.7% 9898.8% 216.7% 157.9% 40.8% 39.4% 38.0% 32.0% 14.7% --
在Python 2.7中:
rate/sec f3 f4 f8 f3b f6 f5 f2 f4b f7 f1
f3 384 -- -2.6% -97.1% -97.2% -97.9% -97.9% -98.0% -98.2% -98.5% -99.2%
f4 394 2.6% -- -97.0% -97.2% -97.8% -97.9% -98.0% -98.1% -98.5% -99.1%
f8 13,079 3303.3% 3216.1% -- -5.6% -28.6% -29.9% -32.8% -38.3% -49.7% -71.2%
f3b 13,852 3504.5% 3412.1% 5.9% -- -24.4% -25.8% -28.9% -34.6% -46.7% -69.5%
f6 18,325 4668.4% 4546.2% 40.1% 32.3% -- -1.8% -5.9% -13.5% -29.5% -59.6%
f5 18,664 4756.5% 4632.0% 42.7% 34.7% 1.8% -- -4.1% -11.9% -28.2% -58.8%
f2 19,470 4966.4% 4836.5% 48.9% 40.6% 6.2% 4.3% -- -8.1% -25.1% -57.1%
f4b 21,187 5413.0% 5271.7% 62.0% 52.9% 15.6% 13.5% 8.8% -- -18.5% -53.3%
f7 26,002 6665.8% 6492.4% 98.8% 87.7% 41.9% 39.3% 33.5% 22.7% -- -42.7%
f1 45,354 11701.5% 11399.0% 246.8% 227.4% 147.5% 143.0% 132.9% 114.1% 74.4% --
你可以看到f1在Python 3.2和2.7下是最快的(或者,更完整地说,在本文顶部的keywithmaxval)
你可以使用运算符。Itemgetter:
import operator
stats = {'a': 1000, 'b': 3000, 'c': 100}
max(stats.iteritems(), key=operator.itemgetter(1))[0]
并且使用stats.iteritems()而不是在内存中构建一个新的列表。max()函数的key参数是一个函数,它计算一个用于确定如何对项目排序的键。
请注意,如果你有另一个键值对'd': 3000,这个方法将只返回其中一个,即使它们都有最大值。
>>> import operator
>>> stats = {'a': 1000, 'b': 3000, 'c': 100, 'd': 3000}
>>> max(stats.iteritems(), key=operator.itemgetter(1))[0]
'b'
如果使用Python3:
>>> max(stats.items(), key=operator.itemgetter(1))[0]
'b'
我测试了接受的答案和@thewolf的最快的解决方案,对一个非常基本的循环,循环比两者都快:
import time
import operator
d = {"a"+str(i): i for i in range(1000000)}
def t1(dct):
mx = float("-inf")
key = None
for k,v in dct.items():
if v > mx:
mx = v
key = k
return key
def t2(dct):
v=list(dct.values())
k=list(dct.keys())
return k[v.index(max(v))]
def t3(dct):
return max(dct.items(),key=operator.itemgetter(1))[0]
start = time.time()
for i in range(25):
m = t1(d)
end = time.time()
print ("Iterating: "+str(end-start))
start = time.time()
for i in range(25):
m = t2(d)
end = time.time()
print ("List creating: "+str(end-start))
start = time.time()
for i in range(25):
m = t3(d)
end = time.time()
print ("Accepted answer: "+str(end-start))
结果:
Iterating: 3.8201940059661865
List creating: 6.928712844848633
Accepted answer: 5.464320182800293