Question

我在R中有一个向量，定义为c(2,3,4,6,5,3,1,3,5,6)。

我想根据向量中所有增加/减少趋势对向量进行子集化。

所需的输出应该是3个子向量（在此示例中）

1) (2,3,4,6)
2) (6,5,3,1)
3) (1,3,5,6)

Answer 1

int main( void )
{
    printf( "Enter a first and last name: " );

    // Keep first letter to output last.
    char letter = getchar() ;

    // Discard the rest of the first name
    while( getchar() != ' ' )
    {
       // Do nothing until SPACE
    }

    // Output the last name
    int namech = 0 ;
    while( (namech = getchar()) != '\n' )
    {
        putchar( namech ) ;
    }
    printf( ", " );

    // Output the first letter
    putchar( letter ) ;
    printf( "." );

    return 0;
}

Answer 2

以下是区分上升趋势和下降趋势的答案

dummy_vector = c(2,3,4,6,5,3,1,3,5,6)

# Loop to mark the trend 
index = rep(1,length(dummy_vector))  # Pre allocate output vector 
for (i in 2:length(dummy_vector)) { 
  if (dummy_vector[i] > dummy_vector[i-1]) {
    index[i] = 1 # trend up
  } 
  else if (dummy_vector[i] < dummy_vector[i-1]) {
    index[i] = 2 # trend down
  } 
  } # end loop

# Mark changes in trend
change_trend = rep(0,length(index))  # Pre allocate output vector 
for (i in 2:length(index)) {
  if (index[i] == 2 && index[i-1] == 1) { 
    change_trend[i-1] = 3
  }
  else if (index[i] == 1 && index[i-1] == 2) { 
    change_trend[i-1] = 3
  }
} # end loop

# Grab index positions 
up_trend_index_start = rep(0,length(index))
up_trend_index_end = rep(0,length(index))
dn_trend_index_start = rep(0,length(index))
dn_trend_index_end = rep(0,length(index))

for (i in 1:length(index)) { 
  if (index[i] == 1 && i == 1) { 
    up_trend_index_start[i] = i
  }
  if (index[i] == 2 && i == 1) { 
    dn_trend_index_start[i] = i
  }
  if (index[i] == 1 && change_trend[i] == 3 ) { 
    up_trend_index_end[i] = i
  }
  if (index[i] == 2 && change_trend[i] == 3 ) { 
    up_trend_index_start[i] = i
  }
  if (index[i] == 1 && change_trend[i] == 3 ) { 
    dn_trend_index_start[i] = i
  }
  if (index[i] == 2 && change_trend[i] == 3 ) { 
    dn_trend_index_end[i] = i
  }
  if (index[i] == 1 && i == length(index)) {
    up_trend_index_end[i] = i
  }
  if (index[i] == 2 && i == length(index)) { 
    dn_trend_index_end[i] = i
  }
  }

# Reduce to remove all 0
up_trend_index_start = up_trend_index_start[up_trend_index_start != 0]
up_trend_index_end = up_trend_index_end[up_trend_index_end != 0]
dn_trend_index_start = dn_trend_index_start[dn_trend_index_start != 0]
dn_trend_index_end = dn_trend_index_end[dn_trend_index_end != 0]

# find maximum vector length
max_i = max(length(up_trend_index_start),length(up_trend_index_end),length(dn_trend_index_start),length(dn_trend_index_end))

# For loop to make subsets 
up_trend = list()
dn_trend = list()
for (i in 1:max_i){ 
  up_trend[[i]] = dummy_vector[up_trend_index_start[i]:up_trend_index_end[i]]
  dn_trend[[i]] = dummy_vector[dn_trend_index_start[i]:dn_trend_index_end[i]]
  if (i >= length(dn_trend_index_end) | i >= length(up_trend_index_end)) { # Break loop if uneven lengths
break
}
}

# Vector output 
up_trend_one = up_trend[[1]]
dn_trend_one = dn_trend[[1]]
up_trend_two = up_trend[[2]]

对于输出

> up_trend_one
[1] 2 3 4 6
> dn_trend_one
[1] 6 5 3 1
> up_trend_two
[1] 1 3 5 6

基于增加/减少趋势的子集向量

2 个答案: