from collections import Counter as cs
l = cs([1, 2, 3, 3, 4])
z = [k if v == 1 else 'a' for k, v in l.items()]
z.remove('a')
print(z)
from collections import Counter as cs
data = 'Python 3 is better than python 2'.split()
count = cs(data)
for i, j in sorted(count.items()):
    print(' ' * (10 - len(i)), end='')
    print('{} :   {}'.format(i, j))
'''
         2 :   1
         3 :   1
    Python :   1
    better :   1
        is :   1
    python :   1
      than :   1
'''
예제 #3
0
from collections import Counter as cs
x = int(input())
y = input().split(" ")
y = cs(y)
z = int(input())
l1, l2 = [], []
for i in range(z):
    q = input().split(" ")
    l1.append(q[0])
    l2.append(q[1])
l3, l4 = [], []
for i in l1:
    if i not in l3:
        l3.append(i)
l5 = []
for i in l3:
    l4 = []
    for j in range(len(l1)):
        if i == l1[j]:
            l4.append(l2[j])
    for j in range(y[i]):
        l5.append(int(l4[j]))
print(sum(l5))
예제 #4
0
l1=[6,2,4,3]
l2=[6,6,1,4,7,1]
l3=[]
for i in l2:
    if i not in l1:
        l1.extend([i]*l2.count(i))
    else:
        if(l1.count(i)<l2.count(i)):
            l1.extend([i]*(l2.count(i)-l1.count(i)))
print(l1)
        
#solution2
from collections import Counter as cs
l1=[6,6,2,4,3]
l2=[6,1,4,7,1]
n1=cs(l1)
n2=cs(l2)
for i,j in n1.items():
    if( i in n2):
        if n2[i]>j:
            print((str(i)+' ')*n2[i],end='')
        else:
            print((str(i)+' ')*j,end='')
    else:
        print((str(i)+' ')*j,end='')
for i,j in n2.items():
    if(i not in n1):
        print((str(i)+' ')*j,end='')
예제 #5
0
from collections import Counter as cs
from collections import defaultdict as dc

x = input().split(" ")
x = list(map(int, x))
for i in range(len(x)):
    l1 = []
    d = dc()
    for j in range(x[i]):
        y = input()
        d[j + 1] = y
        l1.append(y)
    print(l1)
    print(d)
    print(cs(l1))
    a = cs(l1)
    for k, v in a.items():
        if v != 1:
            for i in d.keys():
                if d[i] == k:
                    print(i, end="")
        print()
'''5 2
a
b
a
a
b
['a', 'b', 'a', 'a', 'b']
defaultdict(None, {1: 'a', 2: 'b', 3: 'a', 4: 'a', 5: 'b'})
Counter({'a': 3, 'b': 2})
from collections import Counter as cs,defaultdict as dd,OrderedDict
x=cs(list(input()))
d=dd()
x=(OrderedDict(sorted(x.items(),key=lambda kv:kv[1] ,reverse=True)))
for k,v in x.items():
    if v>1:
        print(k,v)