在Python中按数量对多个列表的元素进行排名
题
我想根据元素在每个列表中出现的频率对多个列表进行排名。例子:
列表 1 = 1,2,3,4
列表2 = 4,5,6,7
列表3 = 4,1,8,9
结果 = 4,1,2,3,4,5,6,7,8(4 算 3 次,1 算 2 次,其余算 1 次)
我已经尝试过以下操作,但我需要一些更智能的东西,以及我可以对任意数量的列表执行的操作。
l = []
l.append([ 1, 2, 3, 4, 5])
l.append([ 1, 9, 3, 4, 5])
l.append([ 1, 10, 8, 4, 5])
l.append([ 1, 12, 13, 7, 5])
l.append([ 1, 14, 13, 13, 6])
x1 = set(l[0]) & set(l[1]) & set(l[2]) & set(l[3])
x2 = set(l[0]) & set(l[1]) & set(l[2]) & set(l[4])
x3 = set(l[0]) & set(l[1]) & set(l[3]) & set(l[4])
x4 = set(l[0]) & set(l[2]) & set(l[3]) & set(l[4])
x5 = set(l[1]) & set(l[2]) & set(l[3]) & set(l[4])
set1 = set(x1) | set(x2) | set(x3) | set(x4) | set(x5)
a1 = list(set(l[0]) & set(l[1]) & set(l[2]) & set(l[3]) & set(l[4]))
a2 = getDifference(list(set1),a1)
print a1
print a2
现在问题来了...我可以用 a3、a4 和 a5 一次又一次地执行此操作,但它太复杂了,我需要一个函数来完成这个...但我不知道如何...我的数学卡住了;)
解决了:非常感谢您的讨论。作为一个新手,我喜欢这个系统:快速+信息丰富。你帮了我所有的忙!泰
解决方案
import collections
data = [
[1, 2, 3, 4, 5],
[1, 9, 3, 4, 5],
[1, 10, 8, 4, 5],
[1, 12, 13, 7, 5],
[1, 14, 13, 13, 6],
]
def sorted_by_count(lists):
counts = collections.defaultdict(int)
for L in lists:
for n in L:
counts[n] += 1
return [num for num, count in
sorted(counts.items(),
key=lambda k_v: (k_v[1], k_v[0]),
reverse=True)]
print sorted_by_count(data)
现在让我们概括它(采用任何可迭代,放宽可散列要求),允许键和反向参数(以匹配排序),然后重命名为 频率排序:
def freq_sorted(iterable, key=None, reverse=False, include_freq=False):
"""Return a list of items from iterable sorted by frequency.
If include_freq, (item, freq) is returned instead of item.
key(item) must be hashable, but items need not be.
*Higher* frequencies are returned first. Within the same frequency group,
items are ordered according to key(item).
"""
if key is None:
key = lambda x: x
key_counts = collections.defaultdict(int)
items = {}
for n in iterable:
k = key(n)
key_counts[k] += 1
items.setdefault(k, n)
if include_freq:
def get_item(k, c):
return items[k], c
else:
def get_item(k, c):
return items[k]
return [get_item(k, c) for k, c in
sorted(key_counts.items(),
key=lambda kc: (-kc[1], kc[0]),
reverse=reverse)]
例子:
>>> import itertools
>>> print freq_sorted(itertools.chain.from_iterable(data))
[1, 5, 4, 13, 3, 2, 6, 7, 8, 9, 10, 12, 14]
>>> print freq_sorted(itertools.chain.from_iterable(data), include_freq=True)
# (slightly reformatted)
[(1, 5),
(5, 4),
(4, 3), (13, 3),
(3, 2),
(2, 1), (6, 1), (7, 1), (8, 1), (9, 1), (10, 1), (12, 1), (14, 1)]
其他提示
结合已经发布的几个想法:
from itertools import chain
from collections import defaultdict
def frequency(*lists):
counter = defaultdict(int)
for x in chain(*lists):
counter[x] += 1
return [key for (key, value) in
sorted(counter.items(), key=lambda kv: (kv[1], kv[0]), reverse=True)]
笔记:
- 在Python 2.7中,您可以使用
Counter
代替defaultdict(int)
. - 该版本采用任意数量的列表作为其参数;前导星号意味着它们将全部打包到一个元组中。如果您想传递包含所有列表的单个列表,请省略该前导星号。
- 如果您的列表包含不可散列的类型,则会中断。
def items_ordered_by_frequency(*lists):
# get a flat list with all the values
biglist = []
for x in lists:
biglist += x
# sort it in reverse order by frequency
return sorted(set(biglist),
key=lambda x: biglist.count(x),
reverse=True)
尝试这一个:
def rank(*lists):
d = dict()
for lst in lists:
for e in lst:
if e in d: d[e] += 1
else: d[e] = 1
return [j[1] for j in sorted([(d[i],i) for i in d], reverse=True)]
用例:
a = [1,2,3,4]
b = [4,5,6,7]
c = [4,1,8,9]
print rank(a,b,c)
可以使用任意数量的列表作为输入
可以计算每个元件(直方图)中出现的次数,再排序它:
def histogram(enumerable):
result = {}
for x in enumerable:
result.setdefault(x, 0)
result[x] += 1
return result
lists = [ [1,2,3,4], [4,5,6,7], ... ]
from itertools import chain
h = histogram(chain(*lists))
ranked = sorted(set(chain(*lists)), key = lambda x : h[x], reverse = True)
试试这个代码:
def elementFreq(myList):
#myList is the list of lists
from collections import Counter
tmp = []
for i in myList: tmp += i
return(Counter(tmp))
请注意:你的名单应该是哈希的类型
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