舉一個if語句在Ruby中

Question

一個變量我有這個if語句：

if value_n_in_f > 0 && value_m_in_f != 1 
    puts "I: f(x)=#{value_m_in_f}x+#{value_n_in_f}" 
    elsif value_n_in_f == 0 && value_m_in_f != 1 
    puts "I: f(x)=#{value_m_in_f}x" 
    elsif value_n_in_f < 0 && value_m_in_f != 1 
    puts "I: f(x)=#{value_m_in_f}x#{value_n_in_f}" 
    elsif value_n_in_f > 0 && value_m_in_f == 1 
    puts "I: f(x)=x+#{value_n_in_f}" 
    elsif value_n_in_f == 0 && value_m_in_f == 1 
    puts "I: f(x)=x" 
    elsif value_n_in_f < 0 && value_m_in_f == 1 
    puts "I: f(x)=x#{value_n_in_f}" 
    end` 

我不得不在其他語句真正經常使用這一說法，它使我的代碼不必要的長時間。很明顯if_for_f =如果.......結束將不起作用。有沒有其他方法？ 這是我希望它看起來像一個例子：

puts "Now we insert #{value_of_x} in #{if_for_f}" 

有什麼辦法，我可以做這樣的事情？請注意，我絕對是這個新手。

由於提前， Kaiton

Answer 1

我不能只是一個高清與它的功能s函數運行這個case 聲明並調用它然後當我需要它時呢？

當然：

def do_stuff(m, n) 
    if m == 1 
    if n > 0  then "I: f(x)=x+#{n}" 
    elsif n == 0 then "I: f(x)=x" 
    elsif n < 0 then "I: f(x)=x#{n}" 
    end 
    else 
    if n > 0  then "I: f(x)=#{m}x+#{n}" 
    elsif n == 0 then "I: f(x)=#{m}x" 
    elsif n < 0 then "I: f(x)=#{m}x#{n}" 
    end 
    end 
end 

puts do_stuff(1, 1) #I: f(x)=x+1 

或者，如果緻密性的目標，我們可以得到這樣的：

def do_stuff(m, n) 
    if m == 1 
    n == 0 ? "I: f(x)=x"  : "I: f(x)=x#{sprintf("%+d", n)}" 
    else 
    n == 0 ? "I: f(x)=#{m}x" : "I: f(x)=#{m}x#{sprintf("%+d", n)}" 
    end 
end 

...然後一個一個襯墊：

def do_stuff(m, n) 
    (m == 1) ? (n == 0 ? "I: f(x)=x" : "I: f(x)=x#{sprintf("%+d", n)}") : (n == 0 ? "I: f(x)=#{m}x" : "I: f(x)=#{m}x#{sprintf("%+d", n)}") 
end 
end 

但是你的方法有零和-1的一些問題：

def do_stuff(value_m_in_f, value_n_in_f) 
if value_n_in_f > 0 && value_m_in_f != 1 
    puts "I: f(x)=#{value_m_in_f}x+#{value_n_in_f}" 
    elsif value_n_in_f == 0 && value_m_in_f != 1 
    puts "I: f(x)=#{value_m_in_f}x" 
    elsif value_n_in_f < 0 && value_m_in_f != 1 
    puts "I: f(x)=#{value_m_in_f}x#{value_n_in_f}" 
    elsif value_n_in_f > 0 && value_m_in_f == 1 
    puts "I: f(x)=x+#{value_n_in_f}" 
    elsif value_n_in_f == 0 && value_m_in_f == 1 
    puts "I: f(x)=x" 
    elsif value_n_in_f < 0 && value_m_in_f == 1 
    puts "I: f(x)=x#{value_n_in_f}" 
    end 
end 

do_stuff(1, 0) 
do_stuff(1,-1) 
do_stuff(1, 1) 
do_stuff(0,-1) 
do_stuff(0, 0) 
do_stuff(-1, 1) 

--output:-- 
I: f(x)=x 
I: f(x)=x-1 
I: f(x)=x+1 
I: f(x)=0x-1 #<---HERE 
I: f(x)=0x  #<---HERE 
I: f(x)=-1x+1 #<---HERE 

所以我們進行了修復：

def get_line_equ(m, b) 
    constant = (b == 0) ? "" : sprintf("%+d", b) # 2 => "+2" 

    case m 
    when 0 
    xterm = "" 
    constant = b 
    when 1 
    xterm = "x" 
    when -1 
    xterm = "-x" 
    else 
    xterm = "#{m}x" 
    end 

    "I: f(x)=#{xterm}#{constant}" 
end 

puts get_line_equ(0, 0) 
puts get_line_equ(0, -1) 
puts get_line_equ(0, 1) 

puts get_line_equ(1, 0) 
puts get_line_equ(1,-1) 
puts get_line_equ(1, 1) 

puts get_line_equ(-1, 0) 
puts get_line_equ(-1, -1) 
puts get_line_equ(-1, 1) 

puts get_line_equ(2, 0) 
puts get_line_equ(2, -1) 
puts get_line_equ(2, 1) 

--output:-- 
I: f(x)=0 
I: f(x)=-1 
I: f(x)=1 
I: f(x)=x 
I: f(x)=x-1 
I: f(x)=x+1 
I: f(x)=-x 
I: f(x)=-x-1 
I: f(x)=-x+1 
I: f(x)=2x 
I: f(x)=2x-1 
I: f(x)=2x+1 

更好？

擾流：

最終DEF是效率不高，因爲它可以是：第一行應被刪除，並複製到每個時分支 - 除了第一。

在迴應評論：

def my_sprintf(str, *numbers) 
    str.gsub(/% .*? [df]/x) do |match| #Looks for %...d or %...f sequences 
    puts match 
    end 
end 

my_sprintf("The answer is: %+d or %+d", -2, 3) 


--output:-- 
%+d 
%+d 

下一頁：

def my_sprintf(str, *numbers) 

    str.gsub(/% .*? [df]/x) do |format_sequ| 
    number_as_str = numbers.shift.to_s 
    p number_as_str 

    if format_sequ[1] == "+" and number_as_str[0] != "-" 
     "+#{number_as_str}" 
    else 
     number_as_str 
    end 

    end 
end 

puts my_sprintf("The answer is: %+d or %+d.", -2, 3) 

--output:-- 
"-2" 
"3" 
The answer is: -2 or +3. 

Answer 2

幾件事情在這裏：你可以把「放」的整個IF/ELSIF塊之前，避免在每一行有看跌期權，就像這樣：

puts case 
    when (value_n_in_f > 0 && value_m_in_f != 1) then "I: f(x)=#{value_m_in_f}x+#{value_n_in_f}" 
    when (value_n_in_f == 0 && value_m_in_f != 1) then "I: f(x)=#{value_m_in_f}x" 
end 

其次，一個case語句會更可讀的，就像這樣：

def big_compare(value_n_in_f, value_m_in_f) 
    msg = case 
    when (value_n_in_f > 0 && value_m_in_f != 1) then "I: f(x)=#{value_m_in_f}x+#{value_n_in_f}" 
    when (value_n_in_f == 0 && value_m_in_f != 1) then "I: f(x)=#{value_m_in_f}x" 
    when (value_n_in_f < 0 && value_m_in_f != 1) then "I: f(x)=#{value_m_in_f}x#{value_n_in_f}" 
    when (value_n_in_f > 0 && value_m_in_f == 1) then "I: f(x)=x+#{value_n_in_f}" 
    when (value_n_in_f == 0 && value_m_in_f == 1) then "I: f(x)=x" 
    when (value_n_in_f < 0 && value_m_in_f == 1) then "I: f(x)=x#{value_n_in_f}" 
    end 
end 
puts big_compare(0, 0) 

Answer 3

這裏是寫你的case語句更緊湊的方式。回想一下a <=> b => -1 if a < b,a <=> b => 0 if a == b和a <=> b => -1 if a > b。

"I: f(x)=" + 
case [value_n_in_f <=> 0, value_m_in_f == 1] 
when [ 1,false] then "#{value_m_in_f}x+#{value_n_in_f}" 
when [ 0,false] then "#{value_m_in_f}x" 
when [-1,false] then "#{value_m_in_f}x#{value_n_in_f}" 
when [ 1,true] then "x+#{value_n_in_f}" 
when [ 0,true] then "x" 
when [-1,true] then "x#{value_n_in_f}" 
end 

如果您希望證明字符串是如何構建的，你可以做這樣的事情（與value_n_in_f和value_m_in_f分別更名爲intercept和slope）：

"I: f(x)=" + 
case 
when slope.zero? 
    intercept.zero? ? "0" : "#{intercept}" 
else 
    case slope.to_f 
    when 1.0 then "" 
    when -1.0 then "-" 
    else "#{slope}" 
    end + "x" + 
    case intercept <=> 0 
    when 0 then "" 
    when -1 then "#{intercept}" 
    else "+#{intercept}"  
    end 
end 

注意，這允許slope < 0，這不是規範的一部分。我測試了這個爲intercept和slope各種組合：

intercept slope string 
     -2.1 4  I: f(x)=4x-2.1 
     -2.1 -2.2 I: f(x)=-2.2x-2.1 
     -2.1 0  I: f(x)=-2.1 
     -2.1 0.0 I: f(x)=-2.1 
     -2.1 -1  I: f(x)=-x-2.1 
     -2.1 1.0 I: f(x)=x-2.1 
     0  4  I: f(x)=4x 
     0  -2.2 I: f(x)=-2.2x 
     0  0  I: f(x)=0 
     0  0.0 I: f(x)=0 
     0  -1  I: f(x)=-x 
     0  1.0 I: f(x)=x 
     0.0 4  I: f(x)=4x 
     0.0 -2.2 I: f(x)=-2.2x 
     0.0 0  I: f(x)=0 
     0.0 0.0 I: f(x)=0 
     0.0 -1  I: f(x)=-x 
     0.0 1.0 I: f(x)=x 
     3  4  I: f(x)=4x+3 
     3  -2.2 I: f(x)=-2.2x+3 
     3  0  I: f(x)=3 
     3  0.0 I: f(x)=3 
     3  -1  I: f(x)=-x+3 
     3  1.0 I: f(x)=x+3