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edited Aug 1, 2018 at 15:57

Python 2, 606606 584 bytes

Saved 22 bytes due to Jo King's golfing.

from itertools import*
s,S,C,E=sum,sorted,combinations,enumerate
def f(a):a=a.split();a.pop(4);e=a.pop(5)if a[-1]=="!"else1]<"$"else 0;b=S("A23456789TJQK".index(i)for i,j in a);d=S(set(b));h=[j for i,j in a];z=len([s(k)for r in range(6)for k in C([10if k+1>10else k+1for[[10,k+1][k<10]for k in b],r)if s(k)==15])*2+s(2for i,j in C(b,2)if i==j)++[4*(0if lene<1),5][len(set(h[:4]h))>1else 5if <2]*(len(set(hh[:4]))<2else 0if e else 4<2)+(a[4][1]in[j for i,j in a[:4]if i=="J"])+s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2]);return z or 19

Try it online!Try it online!

    # import everything from "itertools" library. We only need "combinations" and "groupby".
from itertools import*
# alias functions to shorter names
s,S,C,E=sum,sorted,combinations,enumerate

# function f which takes the hand+up card+crib string as its argument
def f(a):
    # convert space-separated string into list of items.
    a=a.split()
    
    # remove the 4th index, which is always "|".
    a.pop(4) 

    # change golfed by Jo King
    # if the final item in the list is a "!" (if it is <"$"), remove it from the list and assign it to variable "e".
    # otherwise, assign 0 to variable "e".
    # a non-empty string will evaluate to True and 0 will evaluate to False in IF checks later.
    e=a.pop(5)if a[-1]=="!"else1]<"$"else 0
    
    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the value's index in the string "A23456789TJQK".
    # so, ["5S", "5H", "5D", "JS", "KS"] will return [4, 4, 4, 10, 12].
    # using the aliased built-in function sorted(), sort the list numerically ascending.
    b=S("A23456789TJQK".index(i)for i,j in a)
    
    # get the unique items in b, then sort the result numerically ascending.
    d=S(set(b))
    
    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the suits.
    h=[j for i,j in a] 

        # fifteens
        # fifteenschanges golfed by Jo King
        # addgenerate 1pairs toof the(10, value of+ 1) for all cards (since they are zero-indexed)
        # since True and False evaluate to 1 and 0 in python, return 10 if k>=10
        # and reduce all values >10 to 10
        # get all unique combinations of cards for 5 cards, 4 cards, 3 cards, 2 cards, and 1 card
        # add the values of all unique combinations, and return any that equal 15
        # multiply the number of returned 15s by 2 for score
    z=len([s(k)for r in range(6)for k in C([10if k+1>10else k+1for[[10,k+1][k<10]for k in b],r)if s(k)==15])*2
        +
        # pairs
        # using itertools.combinations, get all unique combinations of cards into groups of 2.
        # then, add 2 for each pair where both cards have an identical value.
        s(2for i,j in C(b,2)if i==j)
        +
        # flush
        # usingchanges python'sgolfed ternaryby operatorJo (xKing
 if True else y)    # using list indexing
        # [4 * (0 if thecrib handelse has1), 5], get item at index [0 if more than one suit in hand+up card else 1]
        # else, 5  -> 4 if thenot hand+upcrib cardand hasnot onlyall onesuits suitsame
        # else, 4  -> 5 if theall handcards hassame
 only       #    -> 0 otherwise
        # * (0 if more than one suit in hand else 1)
        #    -> 4 * 0 if not crib and not aall suits same
        #    -> 4 * 1 if not crib handand (alreadyall checkedsuits upsame
 card previously)      #    -> 5 * 1 if all cards same
        # else,   -> 0 otherwise
        [4*(0if lene<1),5][len(set(h[:4]h))>1else 5if <2]*(len(set(hh[:4]))<2else 0if e else 4<2)
        +
        # nobs
        # check if the suit of the 5th card (4, zero-indexed) matches the suit of any of the other 4 cards, and if it does is that card a Jack
        (a[4][1]in[j for i,j in a[:4]if i=="J"])
        +
        # runs
        s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2])
    
    # since only 0 evaluates to false, iff z==0 return 19, else return z.
    returnprint z or 19

Python 2, 606 bytes

from itertools import*
s,S,C,E=sum,sorted,combinations,enumerate
def f(a):a=a.split();a.pop(4);e=a.pop(5)if a[-1]=="!"else 0;b=S("A23456789TJQK".index(i)for i,j in a);d=S(set(b));h=[j for i,j in a];z=len([s(k)for r in range(6)for k in C([10if k+1>10else k+1for k in b],r)if s(k)==15])*2+s(2for i,j in C(b,2)if i==j)+(0if len(set(h[:4]))>1else 5if len(set(h))<2else 0if e else 4)+(a[4][1]in[j for i,j in a[:4]if i=="J"])+s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2]);return z or 19

Try it online!

# import everything from "itertools" library. We only need "combinations" and "groupby".
from itertools import*
# alias functions to shorter names
s,S,C,E=sum,sorted,combinations,enumerate

# function f which takes the hand+up card+crib string as its argument
def f(a):
    # convert space-separated string into list of items.
    a=a.split()
    
    # remove the 4th index, which is always "|".
    a.pop(4)
    
    # if the final item in the list is a "!", remove it from the list and assign it to variable "e".
    # otherwise, assign 0 to variable "e".
    # a non-empty string will evaluate to True and 0 will evaluate to False in IF checks later.
    e=a.pop(5)if a[-1]=="!"else 0
    
    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the value's index in the string "A23456789TJQK".
    # so, ["5S", "5H", "5D", "JS", "KS"] will return [4, 4, 4, 10, 12].
    # using the aliased built-in function sorted(), sort the list numerically ascending.
    b=S("A23456789TJQK".index(i)for i,j in a)
    
    # get the unique items in b, then sort the result numerically ascending.
    d=S(set(b))
    
    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the suits.
    h=[j for i,j in a]
    
        # fifteens
        # add 1 to the value of all cards (since they are zero-indexed), and reduce all values >10 to 10
        # get all unique combinations of cards for 5 cards, 4 cards, 3 cards, 2 cards, and 1 card
        # add the values of all unique combinations, and return any that equal 15
        # multiply the number of returned 15s by 2 for score
    z=len([s(k)for r in range(6)for k in C([10if k+1>10else k+1for k in b],r)if s(k)==15])*2
        +
        # pairs
        # using itertools.combinations, get all unique combinations of cards into groups of 2.
        # then, add 2 for each pair where both cards have an identical value.
        s(2for i,j in C(b,2)if i==j)
        +
        # flush
        # using python's ternary operator (x if True else y)
        # 0 if the hand has more than one suit
        # else, 5 if the hand+up card has only one suit
        # else, 4 if the hand has only one suit and not a crib hand (already checked up card previously)
        # else, 0
        (0if len(set(h[:4]))>1else 5if len(set(h))<2else 0if e else 4)
        +
        # nobs
        # check if the suit of the 5th card (4, zero-indexed) matches the suit of any of the other 4 cards, and if it does is that card a Jack
        (a[4][1]in[j for i,j in a[:4]if i=="J"])
        +
        # runs
        s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2])
    
    # since only 0 evaluates to false, iff z==0 return 19, else return z.
    return z or 19

Python 2, 606 584 bytes

Saved 22 bytes due to Jo King's golfing.

from itertools import*
s,S,C,E=sum,sorted,combinations,enumerate
def f(a):a=a.split();a.pop(4);e=a.pop(5)if a[-1]<"$"else 0;b=S("A23456789TJQK".index(i)for i,j in a);d=S(set(b));h=[j for i,j in a];z=len([s(k)for r in range(6)for k in C([[10,k+1][k<10]for k in b],r)if s(k)==15])*2+s(2for i,j in C(b,2)if i==j)+[4*(e<1),5][len(set(h))<2]*(len(set(h[:4]))<2)+(a[4][1]in[j for i,j in a[:4]if i=="J"])+s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2]);return z or 19

Try it online!

    # import everything from "itertools" library. We only need "combinations" and "groupby".
from itertools import*
# alias functions to shorter names
s,S,C,E=sum,sorted,combinations,enumerate

# function f which takes the hand+up card+crib string as its argument
def f(a):
    # convert space-separated string into list of items.
    a=a.split()

    # remove the 4th index, which is always "|".
    a.pop(4) 

    # change golfed by Jo King
    # if the final item in the list is a "!" (if it is <"$"), remove it from the list and assign it to variable "e".
    # otherwise, assign 0 to variable "e".
    # a non-empty string will evaluate to True and 0 will evaluate to False in IF checks later.
    e=a.pop(5)if a[-1]<"$"else 0

    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the value's index in the string "A23456789TJQK".
    # so, ["5S", "5H", "5D", "JS", "KS"] will return [4, 4, 4, 10, 12].
    # using the aliased built-in function sorted(), sort the list numerically ascending.
    b=S("A23456789TJQK".index(i)for i,j in a)

    # get the unique items in b, then sort the result numerically ascending.
    d=S(set(b))

    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the suits.
    h=[j for i,j in a] 

        # fifteens
        # changes golfed by Jo King
        # generate pairs of (10, value + 1) for all cards (since they are zero-indexed)
        # since True and False evaluate to 1 and 0 in python, return 10 if k>=10
        # and reduce all values >10 to 10
        # get all unique combinations of cards for 5 cards, 4 cards, 3 cards, 2 cards, and 1 card
        # add the values of all unique combinations, and return any that equal 15
        # multiply the number of returned 15s by 2 for score
    z=len([s(k)for r in range(6)for k in C([[10,k+1][k<10]for k in b],r)if s(k)==15])*2
        +
        # pairs
        # using itertools.combinations, get all unique combinations of cards into groups of 2.
        # then, add 2 for each pair where both cards have an identical value.
        s(2for i,j in C(b,2)if i==j)
        +
        # flush
        # changes golfed by Jo King
        # using list indexing
        # [4 * (0 if crib else 1), 5], get item at index [0 if more than one suit in hand+up card else 1]
        #    -> 4 if not crib and not all suits same
        #    -> 5 if all cards same
        #    -> 0 otherwise
        # * (0 if more than one suit in hand else 1)
        #    -> 4 * 0 if not crib and not all suits same
        #    -> 4 * 1 if not crib and all suits same
        #    -> 5 * 1 if all cards same
        #    -> 0 otherwise
        [4*(e<1),5][len(set(h))<2]*(len(set(h[:4]))<2)
        +
        # nobs
        # check if the suit of the 5th card (4, zero-indexed) matches the suit of any of the other 4 cards, and if it does is that card a Jack
        (a[4][1]in[j for i,j in a[:4]if i=="J"])
        +
        # runs
        s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2])

    # since only 0 evaluates to false, iff z==0 return 19, else return z.
    print z or 19

Source Link

created Jul 31, 2018 at 19:04

Triggernometry

Python 2, 606 bytes

from itertools import*
s,S,C,E=sum,sorted,combinations,enumerate
def f(a):a=a.split();a.pop(4);e=a.pop(5)if a[-1]=="!"else 0;b=S("A23456789TJQK".index(i)for i,j in a);d=S(set(b));h=[j for i,j in a];z=len([s(k)for r in range(6)for k in C([10if k+1>10else k+1for k in b],r)if s(k)==15])*2+s(2for i,j in C(b,2)if i==j)+(0if len(set(h[:4]))>1else 5if len(set(h))<2else 0if e else 4)+(a[4][1]in[j for i,j in a[:4]if i=="J"])+s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2]);return z or 19

Try it online!

Slightly shorter than grc's answer, and takes a different route to get there.

Explanation:

# import everything from "itertools" library. We only need "combinations" and "groupby".
from itertools import*
# alias functions to shorter names
s,S,C,E=sum,sorted,combinations,enumerate

# function f which takes the hand+up card+crib string as its argument
def f(a):
    # convert space-separated string into list of items.
    a=a.split()
    
    # remove the 4th index, which is always "|".
    a.pop(4)
    
    # if the final item in the list is a "!", remove it from the list and assign it to variable "e".
    # otherwise, assign 0 to variable "e".
    # a non-empty string will evaluate to True and 0 will evaluate to False in IF checks later.
    e=a.pop(5)if a[-1]=="!"else 0
    
    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the value's index in the string "A23456789TJQK".
    # so, ["5S", "5H", "5D", "JS", "KS"] will return [4, 4, 4, 10, 12].
    # using the aliased built-in function sorted(), sort the list numerically ascending.
    b=S("A23456789TJQK".index(i)for i,j in a)
    
    # get the unique items in b, then sort the result numerically ascending.
    d=S(set(b))
    
    # for each card in the list, split the identifiers into the value(i) and the suit(j).
    # return the suits.
    h=[j for i,j in a]
    
        # fifteens
        # add 1 to the value of all cards (since they are zero-indexed), and reduce all values >10 to 10
        # get all unique combinations of cards for 5 cards, 4 cards, 3 cards, 2 cards, and 1 card
        # add the values of all unique combinations, and return any that equal 15
        # multiply the number of returned 15s by 2 for score
    z=len([s(k)for r in range(6)for k in C([10if k+1>10else k+1for k in b],r)if s(k)==15])*2
        +
        # pairs
        # using itertools.combinations, get all unique combinations of cards into groups of 2.
        # then, add 2 for each pair where both cards have an identical value.
        s(2for i,j in C(b,2)if i==j)
        +
        # flush
        # using python's ternary operator (x if True else y)
        # 0 if the hand has more than one suit
        # else, 5 if the hand+up card has only one suit
        # else, 4 if the hand has only one suit and not a crib hand (already checked up card previously)
        # else, 0
        (0if len(set(h[:4]))>1else 5if len(set(h))<2else 0if e else 4)
        +
        # nobs
        # check if the suit of the 5th card (4, zero-indexed) matches the suit of any of the other 4 cards, and if it does is that card a Jack
        (a[4][1]in[j for i,j in a[:4]if i=="J"])
        +
        # runs
        s(reduce(lambda x,y:x*y,[b.count(k)for k in m])*len(m)for m in[d[s(x[:i]):s(x[:i])+j]for x in[[len(list(e))for i,e in groupby(j-i for i,j in E(d))]]for i,j in E(x)if j>2])
    
    # since only 0 evaluates to false, iff z==0 return 19, else return z.
    return z or 19

Explanation for runs logic specifically:

# for each index and value in the list, add the value minus the index
# since the list is sorted and reduced to unique values, this means adjacent values will all be the same value after offset
# ex: "JD 3C 4H 5H | 5S" -> [2, 3, 4, 10] - > [2, 2, 2, 7]
z = []
for i,j in enumerate(d):
    z.append(j-i)

# group the values by unique value
# then add the length of the groups to the list
# ex: [2, 2, 2, 7] -> [2:[2,2,2], 7:[7]]
#     [2:[2,2,2], 7:[7]] -> [[3], [1]]
w = []
for i,e in groupby(z):
    w.append([len(list(e))])

# list is double-nested so that the combined list comprehension leaves "x" available in both places it is needed
z = []
for x in w:
    for i,j in enumerate(x):
        if j>2:
            # if the group length is larger than 2
            # slice the list of unique card values to obtain only run values
            # since the run can be anywhere in the list, sum the preceding lengths to find the start and end index
            a = d[ sum(x[:i]) : sum(x[:i])+j ]
            z.append(a)

w = []
for m in z:
    # get the number of times the value is in the entire hand
    # ex: "JD 3C 4H 5H | 5S" -> [2,3,4,4,10] and (2,3,4) -> [1, 1, 2]
    a = [b.count(k)for k in m]
    # multiply all values together
    # [1, 1, 2] = 1*1*2 = 2
    a = reduce(lambda x,y:x*y, a)
    # length of the run * number of duplicate values
    a *= len(m)
    w.append(a)

# sum the results of the runs
return sum(w)