Question

我正在研究一种方法，该方法将一个单词数组作为参数并返回一个数组数组，其中每个子数组包含的字是彼此的字谜。第while sort_array[i][1]==temp do行投掷undefined method '[]' for nil:NilClass (NoMethodError)，我不明白为什么。

def combine_anagrams(words)
sort_array = Hash.new

words.each do |w|
    sort_array[w] = w.split("").sort
end

sort_array = sort_array.sort_by { |w, s| s }

return_array = []
i = 0
while i<sort_array.length do
    temp_array = []
    temp = sort_array[i][1]
    while sort_array[i][1]==temp do
        temp_array += [sort_array[i][0]]
        i+=1
    end
    return_array += [temp_array]
end

return temp_array
end

p combine_anagrams( ['cars', 'for', 'potatoes', 'racs', 'four','scar', 'creams','scream'] )

Answer 1

看起来这是因为您正在增加i变量而不检查以确保您仍然在sort_array的范围内。要查看问题，请在代码中最内部puts循环中添加while语句：

while sort_array[i][1]==temp do
  temp_array += [sort_array[i][0]]
  i+=1
  puts "i is #{i} and the length is #{sort_array.length}"
end

然后运行您的代码，您将看到：

i is 1 and the length is 8
i is 2 and the length is 8
i is 3 and the length is 8
i is 4 and the length is 8
i is 5 and the length is 8
i is 6 and the length is 8
i is 7 and the length is 8
i is 8 and the length is 8
example.rb:15:in `combine_anagrams': undefined method `[]' for nil:NilClass (NoMethodError)

您需要确保在两个while循环中保持在数组的范围内，例如：

while i < sort_array.length && sort_array[i][1]==temp do
end

作为旁注，您的代码目前仅返回上一个temp_array，因为您还在外部while循环的开头重置了该代码。而且，如果我了解您尝试解决的问题，可能需要查看阵列上的group_by，这要归功于Enumerable模块：

words = ['cars', 'for', 'potatoes', 'racs', 'four','scar', 'creams','scream']
words.group_by { |word| word.chars.sort }.values
# => [["cars", "racs", "scar"], ["for"], ["potatoes"], ["four"], ["creams", "scream"]]

Ruby：比较数组中的元素时的NoMethodError

1 个答案: