Hướng dẫn ncr python - con trăn ncr
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Bạn có muốn lặp lại? itertools.combination . Cách sử dụng phổ biến: Show
Nội dung chính ShowShow
Nếu bạn chỉ cần tính toán công thức, hãy sử dụng math.factorial : Trong Python 3, sử dụng phép chia số nguyên Đầu ra210 hữu ích 5 bình luận chia sẻ 5 bình luận chia sẻ Do you want iteration? itertools.combinations. Common usage: If you just need to compute the formula, math.factorial can be used, but is not fast for large combinations, but see Output: As of Python 3.8, Do you want iteration? itertools.combinations. Common usage: If you just need to compute the formula, math.factorial can be used, but is not fast for large combinations, but see Output: As of Python 3.8, Do you want iteration? itertools.combinations. Common usage: If you just need to compute the formula, math.factorial can be used, but is not fast for large combinations, but see
If you just need to compute the formula, math.factorial can be used, but is not fast for large combinations, but see Output: As of Python 3.8, Nội dung chính import math def nCr(n,r): f = math.factorial return f(n) // f(r) // f(n-r) if __name__ == '__main__': print nCr(4,2)❮ Math Methods Example Find the total number of possibilities to choose k things from n items: # Import math Libraryimport math # Initialize the number of items to choose fromn = 7 # Initialize the number of possibilities to choosek = 5 # Print total number of possible combinationsprint (math.comb(n, k)) Definition and UsageThe result will be: Run Example » The parameters passed in this method must be positive integers. The >>> import itertools
>>> itertools.combinations('abcd',2)
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| Note: The parameters passed in this method must be positive integers. | Syntax |
|---|---|
| Parameter | Description |
| n | Required. Positive integers of items to choose from |
k If the value of k is greater than the value of n it will return 0 as a result.
Required. Positive integers of items to choose If the parameters are negative, a ValueError occurs. If the parameters are not integers, a TypeError occurs.
Technical Details
| Note: If the value of k is greater than the value of n it will return 0 as a result. | Note: If the parameters are negative, a ValueError occurs. If the parameters are not integers, a TypeError occurs. |
|---|---|
| Return Value: | 3.8 |
Nội dung chính
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
Permutation
❮ Math Methods
Python3
Example
Find the total number of possibilities to choose k things from n items:
# Import math Libraryimport math
# Initialize the number of items to choose fromn = 7
Output:
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
# Initialize the number of possibilities to choosek = 5
If want to get permutations of length L then implement it in this way.
Python3
Example
Find the total number of possibilities to choose k things from n items:
# Import math Libraryimport math
# Initialize the number of items to choose fromn = 7
Output:
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
# Initialize the number of possibilities to choosek = 5
Combination
# Print total number of possible combinationsprint (math.comb(n, k))
Python3
The result will be:
Run Example »
The
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
Note: The parameters passed in this method must be positive integers.
Output:
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
Syntax
Python3
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import math
>>> math.comb(4,2)
6
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
6
6
6
Output:
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
2. Các yếu tố được coi là duy nhất dựa trên vị trí của chúng, chứ không phải dựa trên giá trị của chúng. Vì vậy, nếu các phần tử đầu vào là duy nhất, sẽ không có giá trị lặp lại trong mỗi kết hợp. & Nbsp; & nbsp;
Python3
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import math
>>> math.comb(4,2)
6
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
6
6
6
Output:
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
2. Các yếu tố được coi là duy nhất dựa trên vị trí của chúng, chứ không phải dựa trên giá trị của chúng. Vì vậy, nếu các phần tử đầu vào là duy nhất, sẽ không có giá trị lặp lại trong mỗi kết hợp. & Nbsp; & nbsp;
Python3
Python3
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import math
>>> math.comb(4,2)
6
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
6
6
6
Output:
2. Các yếu tố được coi là duy nhất dựa trên vị trí của chúng, chứ không phải dựa trên giá trị của chúng. Vì vậy, nếu các phần tử đầu vào là duy nhất, sẽ không có giá trị lặp lại trong mỗi kết hợp. & Nbsp; & nbsp;How do you calculate the number of possible combinations?
Python3nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time. To calculate a combination, you will need to calculate a factorial.
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
1>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
8 import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
3import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
2import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
1import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
0import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
1import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
4>>> import math
>>> math.comb(4,2)
6
6import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
22
3. Nếu chúng ta muốn tạo kết hợp cùng một yếu tố với cùng một phần tử thì chúng ta sử dụng Combinations_with_Replocation..
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import itertools
>>> itertools.combinations('abcd',2)
>>> list(itertools.combinations('abcd',2))
[('a', 'b'), ('a', 'c'), ('a', 'd'), ('b', 'c'), ('b', 'd'), ('c', 'd')]
>>> [''.join(x) for x in itertools.combinations('abcd',2)]
['ab', 'ac', 'ad', 'bc', 'bd', 'cd']
import math
def nCr(n,r):
f = math.factorial
return f(n) / f(r) / f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
>>> import math
>>> math.comb(4,2)
6
import math
def nCr(n,r):
f = math.factorial
return f(n) // f(r) // f(n-r)
if __name__ == '__main__':
print nCr(4,2)
6
6
6
6
Để tính toán các kết hợp, chúng tôi sẽ sử dụng công thức ncr = n! / r!* (n - r) !, trong đó n đại diện cho tổng số mặt hàng và r đại diện cho số lượng vật phẩm được chọn tại một thời điểm. Để tính toán một sự kết hợp, bạn sẽ cần phải tính toán một giai thừa.
Làm thế nào để bạn tìm thấy sự kết hợp của 4 số trong Python?
Tất cả các kết hợp có thể có của 4 số.
