JavaScript 排名排行榜
JavaScript Rank Leaderboard
[
[ 'UserA', 490 ],
[ 'UserC', 175 ],
[ 'UserD', 67 ],
[ 'UserB', 26 ]
]
我如何根据排行榜给他们排名?例如。 UserA 排名第一,UserC 排名第二,依此类推
它将看起来像这样:
UserA: Rank 1
UserC: Rank 2
UserD: Rank 3
UserB: Rank 4
使用对象并使用排序函数对其进行排序
const results = {
'UserA': { points: 490 },
'UserB': { points: 26 },
'UserC': { points: 175 },
'UserD': { points: 67 }
}
const rank = Object.entries(results)
.sort(([, a], [, b]) => b.points - a.points)
.map((item, i) => {
results[item[0]].rank = (i + 1)
console.log(item[0], results[item[0]].rank)
return `${item[0]}'s Rank: ${i+1}`
})
console.log(results)
const findUserRank = user => rank.filter(entry => entry.includes(user))
const getUserRankOnly = user => results[user].rank
const getRankUser = ranking => Object.entries(results).filter(([key,val]) => val.rank===ranking).map(([key]) => key)
console.log("Users containing 'User':", findUserRank('User').join("\n"))
console.log("User with rank #3", getRankUser(3)[0])
console.log("UserC's rank:", getUserRankOnly('UserC'))
与嵌套数组相同
const results = [
[ 'UserA', 490 ],
[ 'UserC', 175 ],
[ 'UserD', 67 ],
[ 'UserB', 26 ]
];
const rank = results.slice(0)
.sort(([, a], [, b]) => b - a)
.map((item,i) => `${item[0]}'s Rank: ${i+1}`);
const getUserRank = user => rank.filter(entry => entry.includes(user))
const getUserRankOnly = user => rank.filter(entry => entry.includes(user))[0].split(": ")[1]
const getRankUser = ranking => rank.filter(entry => entry.endsWith(ranking))[0].split("'")[0]
console.log("Users containing 'User':",getUserRank('User').join("\n"))
console.log("Users containing 'UserC':",getUserRank('UserC')[0])
console.log("User with rank #3",getRankUser('3'))
console.log("UserC's rank:",getUserRankOnly('UserC')[0])
[
[ 'UserA', 490 ],
[ 'UserC', 175 ],
[ 'UserD', 67 ],
[ 'UserB', 26 ]
]
我如何根据排行榜给他们排名?例如。 UserA 排名第一,UserC 排名第二,依此类推
它将看起来像这样:
UserA: Rank 1
UserC: Rank 2
UserD: Rank 3
UserB: Rank 4
使用对象并使用排序函数对其进行排序
const results = {
'UserA': { points: 490 },
'UserB': { points: 26 },
'UserC': { points: 175 },
'UserD': { points: 67 }
}
const rank = Object.entries(results)
.sort(([, a], [, b]) => b.points - a.points)
.map((item, i) => {
results[item[0]].rank = (i + 1)
console.log(item[0], results[item[0]].rank)
return `${item[0]}'s Rank: ${i+1}`
})
console.log(results)
const findUserRank = user => rank.filter(entry => entry.includes(user))
const getUserRankOnly = user => results[user].rank
const getRankUser = ranking => Object.entries(results).filter(([key,val]) => val.rank===ranking).map(([key]) => key)
console.log("Users containing 'User':", findUserRank('User').join("\n"))
console.log("User with rank #3", getRankUser(3)[0])
console.log("UserC's rank:", getUserRankOnly('UserC'))
与嵌套数组相同
const results = [
[ 'UserA', 490 ],
[ 'UserC', 175 ],
[ 'UserD', 67 ],
[ 'UserB', 26 ]
];
const rank = results.slice(0)
.sort(([, a], [, b]) => b - a)
.map((item,i) => `${item[0]}'s Rank: ${i+1}`);
const getUserRank = user => rank.filter(entry => entry.includes(user))
const getUserRankOnly = user => rank.filter(entry => entry.includes(user))[0].split(": ")[1]
const getRankUser = ranking => rank.filter(entry => entry.endsWith(ranking))[0].split("'")[0]
console.log("Users containing 'User':",getUserRank('User').join("\n"))
console.log("Users containing 'UserC':",getUserRank('UserC')[0])
console.log("User with rank #3",getRankUser('3'))
console.log("UserC's rank:",getUserRankOnly('UserC')[0])