Question

我有一个包含重复元素的列表。我想计算每个元素的出现次数

示例 - ＆GT; list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

输出 - ＆GT; result = ["a1", "b2", "c1", "d3", "a2"]

即。输出格式应为<value><count>

我尝试过这个解决方案，但它没有给出预期的输出

def find_repetitive_elem(a):
    dict_obj = {}
    arr_obj = []
    for i in range(0, len(a)-1):
        count = 1    
        if a[i] == a[i+1]
            set_val = "%s%s" % (a[i], count+1)
        else:
            set_val = "%s%s" % (a[i], count)                
        arr_obj.append(set_val)

我正在尝试使用算法而不是python包来实现它！

Answer 1

您可以使用itertools.groupby：

from itertools import groupby

list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

answer = [group[0] + str(len(list(group[1]))) for group in groupby(list1)]
print(answer)

<强>输出

['a1', 'b2', 'c1', 'd3', 'a2']

Answer 2

您可以使用groupby模块中的itertools来实现此目的：

import itertools
list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]
["{}{}".format(key,sum( 1 for _ in group )) for key, group in itertools.groupby( list1 ) ]

当您更新了答案，表明您不想使用Python软件包时，这是我可以想到的最优雅的方式，不使用外部模块：

list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

#output-> result = ["a1", "b2", "c1", "d3", "a2"]

def find_repetitive_elem(array):
    output=[]           
    for letter in array:
        existing_entry = {}
        new_entry = {}
        if len(output)>0:            
            existing_entry= output.pop();            
        if letter in existing_entry:            
            existing_entry[letter] +=1            
            output.append(existing_entry)
        else:            
            if existing_entry:            
                output.append(existing_entry)
            new_entry[letter]=1
            output.append(new_entry)                        

    return output

output = find_repetitive_elem(list1)        
print ["{}{}".format(key,value) for key,value in (entry.popitem() for entry in output)]

Answer 3

不确定你的意思

我正在尝试使用算法而不是python来实现它包的！

这不使用itertools（尽管这是一种更好的方法）

def compact(inval):
    last_char = None
    outval = []
    count = 0
    for l in inval + ['dummyvalue']:
        if last_char and last_char != l:
            outval.append("{}{}".format(last_char, count))
            count = 0
        last_char = l
        count += 1
    return outval

letters = ["a", "b", "b", "c", "d", "d", "d", "a", "a"]
print(compact(letters))

输出

letters = ["a", "b", "b", "c", "d", "d", "d", "a", "a"] # yields
['a1', 'b2', 'c1', 'd3', 'a2']

letters = ["a", "b", "b", "c", "c", "d", "e", "d", "d", "d", "a", "a"] # yields
['a1', 'b2', 'c2', 'd1', 'e1', 'd3', 'a2']

Answer 4

由于您不想使用任何Python的模块，因此这是一个不使用任何模块的解决方案（隐式地除了builtins）。代码应该能够处理的不仅仅是字符串。

def main():
    array = 'a', 'b', 'b', 'c', 'd', 'd', 'd', 'a', 'a'
    result = find_repetitive_elements(array)
    print(result)


def find_repetitive_elements(array):
    result, first, last, count = [], True, None, None
    for item in array:
        if first:
            first, last, count = False, item, 1
        elif item == last:
            count += 1
        else:
            result.append('{}{}'.format(last, count))
            last, count = item, 1
    if not first:
        result.append('{}{}'.format(last, count))
    return result


if __name__ == '__main__':
    main()

Answer 5

这可能看起来很基本，但这就是我理解这个问题的方法：

<强>代码：

list1 = ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

def find_repetitive_elem_continuously(arr):
    result = list()
    prev_ele = arr[0]
    count = 0
    for curr_ele in arr:
        if curr_ele != prev_ele:
            new_arr_ele = str(prev_ele)+""+str(count)
            result.append(new_arr_ele)
            prev_ele = curr_ele
            count = 1
        else:
            count += 1
    new_arr_ele = str(prev_ele)+""+str(count)
    result.append(new_arr_ele)
    print result

find_repetitive_elem_continuously(list1)

<强>输出：

['a1', 'b2', 'c1', 'd3', 'a2']

Answer 6

这是我未使用itertools.groupby的尝试。代码依赖于itertools.dropwhile迭代器，但是替换它很容易，我把它留作练习。

import itertools

list1= ["a", "b", "b", "c", "d", "d", "d", "a", "a"]

result = []
lst = list1.copy()

while len(lst)>0:
    current_char = lst[0]
    length = len(lst)
    lst = list(itertools.dropwhile(lambda x: x==current_char, lst))
    result.append("{}{}".format(current_char, length-len(lst)))

print(result)

请参阅http://ideone.com/Cd1Y8B

的代码输出

在列表中如何计算序列中的字符串

6 个答案: