查找元组列表中具有相同第一项和第三项的元组的计数
Finding count of tuples with same first and third item in list of tuples
我有一个元组列表，每个元组包含三个项目：
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
我想在列表中查找第一项和第三项相同的元组的数量，例如第一项 1 和第三项 2015，有 4 个元组；第一项 2 和第三项 2015，有 4 个元组。
我试过了：
for tup in z:
a=tup[0]
b=tup[2]
print(len(set({a:b})))
它没有给出预期的结果。怎么做？
在纯 python 中使用 Counter 和生成器，谢谢@Felix:
from collections import Counter
out = Counter((x[0], x[2]) for x in z)
print (out)
Counter({(3, 2015): 5,
(5, 2015): 5,
(8, 2015): 5,
(11, 2015): 5,
(1, 2016): 5,
(1, 2015): 4,
(2, 2015): 4,
(4, 2015): 4,
(6, 2015): 4,
(7, 2015): 4,
(9, 2015): 4,
(10, 2015): 4,
(12, 2015): 4})
在 pandas 中按 GroupBy.size 聚合计数，输出为 Series:
s = pd.DataFrame(z).groupby([0,2]).size()
print (s)
0 2
1 2015 4
2016 5
2 2015 4
3 2015 5
4 2015 4
5 2015 5
6 2015 4
7 2015 4
8 2015 5
9 2015 4
10 2015 4
11 2015 5
12 2015 4
dtype: int64
使用collections.
例如：
import collections
d = collections.defaultdict(int)
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
for i in z:
d[(i[0], i[2])] += 1
print(d)
输出：
defaultdict(<type 'int'>, {(10, 2015): 4, (5, 2015): 5, (2, 2015): 4, (11, 2015): 5, (6, 2015): 4, (8, 2015): 5, (3, 2015): 5, (12, 2015): 4, (7, 2015): 4, (9, 2015): 4, (4, 2015): 4, (1, 2016): 5, (1, 2015): 4})
使用标准 python 的 itertools.groupby:
from itertools import groupby
for grp, elmts in groupby(z, lambda x: (x[0], x[2])):
print(grp, len(list(elmts)))
编辑：
一个更好的解决方案，使用 operator.itemgetter 而不是 lambda：
from operator import itemgetter
from itertools import groupby
for grp, elmts in groupby(z, itemgetter(0, 2)):
print(grp, len(list(elmts)))
输出：
(1, 2015) 4
(2, 2015) 4
(3, 2015) 5
(4, 2015) 4
(5, 2015) 5
(6, 2015) 4
(7, 2015) 4
(8, 2015) 5
(9, 2015) 4
(10, 2015) 4
(11, 2015) 5
(12, 2015) 4
(1, 2016) 5
将 collections.Counter 与 operator.itemgetter 一起使用：
from collections import Counter
from operator import itemgetter
res = Counter(map(itemgetter(0, 2), z))
print(res)
Counter({(1, 2015): 4,
(1, 2016): 5,
(2, 2015): 4,
(3, 2015): 5,
(4, 2015): 4,
(5, 2015): 5,
(6, 2015): 4,
(7, 2015): 4,
(8, 2015): 5,
(9, 2015): 4,
(10, 2015): 4,
(11, 2015): 5,
(12, 2015): 4})
您可以将计数存储在一个字典中，由一个元组作为键，该元组由原始元组列表中的第一项和第三项组成，例如：
import collections
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015),
(2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015),
(3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015),
(5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015),
(6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015),
(8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015),
(9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015),
(10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015),
(11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015),
(12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
counter = collections.defaultdict(int) # Use a dict factory to save some time
for element in z: # iterate over the tuples
counter[(element[0], element[2])] += 1 # increase the count for each match
# finally, lets print the results
for k, count in counter.items():
print("{}: {}".format(k, count))
哪个会给你：
(1, 2015): 4
(2, 2015): 4
(3, 2015): 5
(4, 2015): 4
(5, 2015): 5
(6, 2015): 4
(7, 2015): 4
(8, 2015): 5
(9, 2015): 4
(10, 2015): 4
(11, 2015): 5
(12, 2015): 4
(1, 2016): 5
from collections import Counter
tmp = [(x[0],x[2]) for x in z]
print(Counter(tmp))
输出会像
Counter({(5, 2015): 5, (11, 2015): 5, (8, 2015): 5, (3, 2015): 5, (1, 2016): 5, (10, 2015): 4, (2, 2015): 4, (6, 2015): 4, (12, 2015): 4, (7, 2015): 4, (9, 2015): 4, (4, 2015): 4, (1, 2015): 4})
试试这个：
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
newz = [(i[0],i[-1]) for i in z]
for i in list(set(newz)):
print(str(i)+' '+str(newz.count(i)))
输出：
(10, 2015) 4
(5, 2015) 5
(2, 2015) 4
(11, 2015) 5
(6, 2015) 4
(8, 2015) 5
(3, 2015) 5
(12, 2015) 4
(7, 2015) 4
(9, 2015) 4
(1, 2016) 5
(4, 2015) 4
(1, 2015) 4
groupby以外的解决方法，
import pprint
import random
from collections import Counter
z = [] # creating random dates as user has 2 years, won't work if year range increases
num_dates = 20
counts_by_month_and_year = Counter()
while len(z) < num_dates:
new = (random.randrange(1, 31), random.randrange(1, 12), random.randrange(2015, 2016))
z.append(new)
counts_by_month_and_year[(new[0], new[2])] += 1
pprint.pprint(dict(counts_by_month_and_year)) # formatting the output
{(1, 2015): 1,
(3, 2015): 1,
(4, 2015): 1,
(5, 2015): 1,
(7, 2015): 1,
(8, 2015): 2,
(9, 2015): 1,
(11, 2015): 1,
(13, 2015): 1,
(16, 2015): 1,
(17, 2015): 1,
(20, 2015): 1,
(21, 2015): 2,
(22, 2015): 1,
(25, 2015): 1,
(26, 2015): 1,
(27, 2015): 2}
[Program finished]
我有一个元组列表，每个元组包含三个项目：
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
我想在列表中查找第一项和第三项相同的元组的数量，例如第一项 1 和第三项 2015，有 4 个元组；第一项 2 和第三项 2015，有 4 个元组。
我试过了：
for tup in z:
a=tup[0]
b=tup[2]
print(len(set({a:b})))
它没有给出预期的结果。怎么做？
在纯 python 中使用 Counter 和生成器，谢谢@Felix:
from collections import Counter
out = Counter((x[0], x[2]) for x in z)
print (out)
Counter({(3, 2015): 5,
(5, 2015): 5,
(8, 2015): 5,
(11, 2015): 5,
(1, 2016): 5,
(1, 2015): 4,
(2, 2015): 4,
(4, 2015): 4,
(6, 2015): 4,
(7, 2015): 4,
(9, 2015): 4,
(10, 2015): 4,
(12, 2015): 4})
在 pandas 中按 GroupBy.size 聚合计数，输出为 Series:
s = pd.DataFrame(z).groupby([0,2]).size()
print (s)
0 2
1 2015 4
2016 5
2 2015 4
3 2015 5
4 2015 4
5 2015 5
6 2015 4
7 2015 4
8 2015 5
9 2015 4
10 2015 4
11 2015 5
12 2015 4
dtype: int64
使用collections.
例如：
import collections
d = collections.defaultdict(int)
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
for i in z:
d[(i[0], i[2])] += 1
print(d)
输出：
defaultdict(<type 'int'>, {(10, 2015): 4, (5, 2015): 5, (2, 2015): 4, (11, 2015): 5, (6, 2015): 4, (8, 2015): 5, (3, 2015): 5, (12, 2015): 4, (7, 2015): 4, (9, 2015): 4, (4, 2015): 4, (1, 2016): 5, (1, 2015): 4})
使用标准 python 的 itertools.groupby:
from itertools import groupby
for grp, elmts in groupby(z, lambda x: (x[0], x[2])):
print(grp, len(list(elmts)))
编辑：
一个更好的解决方案，使用 operator.itemgetter 而不是 lambda：
from operator import itemgetter
from itertools import groupby
for grp, elmts in groupby(z, itemgetter(0, 2)):
print(grp, len(list(elmts)))
输出：
(1, 2015) 4
(2, 2015) 4
(3, 2015) 5
(4, 2015) 4
(5, 2015) 5
(6, 2015) 4
(7, 2015) 4
(8, 2015) 5
(9, 2015) 4
(10, 2015) 4
(11, 2015) 5
(12, 2015) 4
(1, 2016) 5
将 collections.Counter 与 operator.itemgetter 一起使用：
from collections import Counter
from operator import itemgetter
res = Counter(map(itemgetter(0, 2), z))
print(res)
Counter({(1, 2015): 4,
(1, 2016): 5,
(2, 2015): 4,
(3, 2015): 5,
(4, 2015): 4,
(5, 2015): 5,
(6, 2015): 4,
(7, 2015): 4,
(8, 2015): 5,
(9, 2015): 4,
(10, 2015): 4,
(11, 2015): 5,
(12, 2015): 4})
您可以将计数存储在一个字典中，由一个元组作为键，该元组由原始元组列表中的第一项和第三项组成，例如：
import collections
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015),
(2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015),
(3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015),
(5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015),
(6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015),
(8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015),
(9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015),
(10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015),
(11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015),
(12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
counter = collections.defaultdict(int) # Use a dict factory to save some time
for element in z: # iterate over the tuples
counter[(element[0], element[2])] += 1 # increase the count for each match
# finally, lets print the results
for k, count in counter.items():
print("{}: {}".format(k, count))
哪个会给你：
(1, 2015): 4 (2, 2015): 4 (3, 2015): 5 (4, 2015): 4 (5, 2015): 5 (6, 2015): 4 (7, 2015): 4 (8, 2015): 5 (9, 2015): 4 (10, 2015): 4 (11, 2015): 5 (12, 2015): 4 (1, 2016): 5
from collections import Counter
tmp = [(x[0],x[2]) for x in z]
print(Counter(tmp))
输出会像
Counter({(5, 2015): 5, (11, 2015): 5, (8, 2015): 5, (3, 2015): 5, (1, 2016): 5, (10, 2015): 4, (2, 2015): 4, (6, 2015): 4, (12, 2015): 4, (7, 2015): 4, (9, 2015): 4, (4, 2015): 4, (1, 2015): 4})
试试这个：
z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]
newz = [(i[0],i[-1]) for i in z]
for i in list(set(newz)):
print(str(i)+' '+str(newz.count(i)))
输出：
(10, 2015) 4
(5, 2015) 5
(2, 2015) 4
(11, 2015) 5
(6, 2015) 4
(8, 2015) 5
(3, 2015) 5
(12, 2015) 4
(7, 2015) 4
(9, 2015) 4
(1, 2016) 5
(4, 2015) 4
(1, 2015) 4
groupby以外的解决方法，
import pprint
import random
from collections import Counter
z = [] # creating random dates as user has 2 years, won't work if year range increases
num_dates = 20
counts_by_month_and_year = Counter()
while len(z) < num_dates:
new = (random.randrange(1, 31), random.randrange(1, 12), random.randrange(2015, 2016))
z.append(new)
counts_by_month_and_year[(new[0], new[2])] += 1
pprint.pprint(dict(counts_by_month_and_year)) # formatting the output
{(1, 2015): 1,
(3, 2015): 1,
(4, 2015): 1,
(5, 2015): 1,
(7, 2015): 1,
(8, 2015): 2,
(9, 2015): 1,
(11, 2015): 1,
(13, 2015): 1,
(16, 2015): 1,
(17, 2015): 1,
(20, 2015): 1,
(21, 2015): 2,
(22, 2015): 1,
(25, 2015): 1,
(26, 2015): 1,
(27, 2015): 2}
[Program finished]