Question

我具有任何输入返回字符串black的功能。我想将列表中不等于yellow的每个元素都更改为黑色，并将其作为新列表返回。

使用过滤器并将其映射会从列表中删除我不想做的元素。

我也想使用函数式编程来保留它，lambdas对于循环来说并不老。我知道我可以在map函数中指定过滤条件，但我想避免这种情况

def change_color(color:String):String = {
return "black"
}

val cats = Seq("pink","blue","yellow")

所以输出应该是

Seq("black","black","yellow")

Answer 1

这是映射函数在集合上的常见应用。您的情况是一个列表。地图的本质是什么？

def change_color(color: String) : String =  
   if (color == "yellow") "yellow" else "black"


def map[A, B](list: List[A)(f: A => B): List[B] = {
   ???
}

它将函数从A到B应用于集合的每个元素，因此，在您的情况下，您的函数为String => String

cats.map(change_color _)

cats.map(el => change_color(el))

这只是使用Scala集合及其功能的开始。从这里您可以尝试折叠，平面图，分组，并行处理等。

Answer 2

或者，您可以使用protected function _login( $username, $password, $forceLogin = false, $appRefreshInterval = 1800) { if (empty($username) || empty($password)) { throw new \InvalidArgumentException('You must provide a username and password to _login().'); } // Switch the currently active user/pass if the details are different. if ($this->username !== $username || $this->password !== $password) { $this->_setUser($username, $password); } // Perform a full relogin if necessary. if (!$this->isMaybeLoggedIn || $forceLogin) { $this->_sendPreLoginFlow(); try { $response = $this->request('accounts/login/') ->setNeedsAuth(false) ->addPost('phone_id', $this->phone_id) ->addPost('_csrftoken', $this->client->getToken()) ->addPost('username', $this->username) ->addPost('adid', $this->advertising_id) ->addPost('guid', $this->uuid) ->addPost('device_id', $this->device_id) ->addPost('password', $this->password) ->addPost('login_attempt_count', 0) ->getResponse(new Response\LoginResponse()); } catch (\InstagramAPI\Exception\InstagramException $e) { if ($e->hasResponse() && $e->getResponse()->isTwoFactorRequired()) { // Login failed because two-factor login is required. // Return server response to tell user they need 2-factor. return $e->getResponse(); } else { // Login failed for some other reason... Re-throw error. throw $e; } } $this->_updateLoginState($response); $this->_sendLoginFlow(true, $appRefreshInterval); // Full (re-)login successfully completed. Return server response. return $response; } // Attempt to resume an existing session, or full re-login if necessary. // NOTE: The "return" here gives a LoginResponse in case of re-login. return $this->_sendLoginFlow(false, $appRefreshInterval); }来转换此序列

Partial Function

这将更加简洁。

Scala-如何仅对列表中的某些元素进行映射而无需过滤器？

2 个答案: