Question

假设：

a = [[1,2,3,4],
     [1,2,3,7],
     [1,2,3,4]]

输出7的位置为（1,3）需要做什么？

我尝试使用.index无济于事。

Answer 1

require 'matrix'

a = [[1, 2, 3, 4],
     [1, 2, 3, 7],
     [1, 2, 3, 4]]

Matrix[*a].index(7)

=> [1, 3]

Answer 2

如果子阵列的宽度相同，则可以将其展平为单个阵列，并将位置视为row_num * row_width + col_num：

idx = a.flatten.index(7)
row_width = a[0].length
row = idx / row_width
col = idx - (row * row_width)
puts [row, col] # => [1, 3]

或者您可以迭代它以查找所有匹配项：

def find_indices_for(array, value)
  array.with_object([]).with_index do |(row, matches), row_index|
    matches << [row_index, row.index(value)] if row.index(value)
  end
end

find_indices_for(a, 7) # => [[1, 3]]
find_indices_for(a, 2) # => [[0, 1], [1, 1], [2, 1]]

Answer 3

<int-kafka:message-driven-channel-adapter>在这里效果很好：

KafkaMessageHeaders kafkaMessageHeaders = new KafkaMessageHeaders(generateMessageId, generateTimestamp);

Map<String, Object> rawHeaders = kafkaMessageHeaders.getRawHeaders();
rawHeaders.put(KafkaHeaders.MESSAGE_KEY, key);
rawHeaders.put(KafkaHeaders.TOPIC, metadata.getPartition().getTopic());
rawHeaders.put(KafkaHeaders.PARTITION_ID, metadata.getPartition().getId());
rawHeaders.put(KafkaHeaders.OFFSET, metadata.getOffset());
rawHeaders.put(KafkaHeaders.NEXT_OFFSET, metadata.getNextOffset());

if (!this.autoCommitOffset) {
    rawHeaders.put(KafkaHeaders.ACKNOWLEDGMENT, acknowledgment);
}

现在，each_with_index将返回一个数组，其中包含def locator(array, number) locations = Array.new array.each_with_index do |mini_array, index| mini_array.each_with_index do |element, sub_index| locations << [index, sub_index] if element == number end end locations end中locator(array, number)的所有位置。

Answer 4

def locs2D(a,e)
  a.size.times.with_object([]) do |row,arr|
    row.size.times { |col| arr << [row,col] if a[row][col] == e }
  end
end

locs2D(a,7) #=> [[1, 3]] 
locs2D(a,3) #=> [[0, 2], [1, 2], [2, 2]]

如何在ruby中的数组数组中找到整数的位置？

4 个答案: