Question

我有一个列表列表：

ls = [['01',2,3,4], ['02',5,2,4], ['03',2,6,4], ['01',1,3,4]]

我想在该字符串之后添加该子列表的术语，并从相同的字符串开始的所有其他子列表中添加相应的术语。

所以结果应该像这样：

result = [['01',3,6,8], ['02',5,2,4], ['03',2,6,4]]
# the first and last original sublists were "combined"

我尝试了以下代码：

ls=[['01',2,3,4],['02',5,2,4],['03',2,6,4],['01',1,3,4]]
totals = {}
for key, value1,value2,value3 in ls:
    totals[key] = totals.get(key, 0) + value1
    totals[key] = totals.get(key, 0) + value2
    totals[key] = totals.get(key, 0) + value3

print(totals)

但这不是我的目标：

{'01': 17, '02': 11, '03': 12}

Answer 1

对您的代码进行少量修改，当有新的“键”时，我已经初始化了值[0, 0, 0]。

ls = [['01', 2, 3, 4],['02', 5, 2, 4],['03', 2, 6, 4],['01', 1, 3, 4]]
totals = {}
for key, value1, value2, value3 in ls:
    if key not in totals.keys():
        totals[key] = [0, 0, 0]
    totals[key][0] += value1
    totals[key][1] += value2
    totals[key][2] += value3

totals = [[k] + v for k, v in totals.items()]
print(totals)

# Output:
# [['01', 3, 6, 8], ['02', 5, 2, 4], ['03', 2, 6, 4]]

您的代码的第一个问题是，当您拥有新密钥时，totals.get(key, 0)不会给您0我们想要的初始值[0, 0, 0]。

第二，您要在for循环中对列表的所有值求和，而不是逐个元素地对列表求和。因此，您得到{'01': 17, '02': 11, '03': 12}

版本2（更通用，可处理非数字）

ls = [['01', 2, 'a', 4],['02', 5, 2, 4],['03', 2, 6, 4],['01', 1, 'ad', 4]]
totals = {}
for x in ls:
    if x[0] not in totals.keys():
        totals[x[0]] = x[1:]
    else:
        for i in range(len(totals[x[0]])):
            if isinstance(x[1 + i], (int, float)):
                totals[x[0]][i] += x[1 + i]

totals = [[k] + v for k, v in totals.items()]
print(totals)

# Output:
# [['01', 3, 'a', 8], ['02', 5, 2, 4], ['03', 2, 6, 4]]

Answer 2

您可以使用setdefault以及一些索引来获得所需的结果。

.well-known

编辑以处理出现字符而选择第一个字符的情况

ls=[['01',2,3,4],['02',5,2,4],['03',2,6,4],['01',1,3,4]]
totals = {}
for element in ls:
    key ,array = element[0],element[1:]
    for index in range(len(array)):
        totals.setdefault(key,[0]*len(array))[index] += array[index]    
result = []
for key,value in totals.items():
    result.append([key] + value)
print(result)

# you can sort the result to get exactly as you wanted
# you dont need this step if the dictionary is already sorted like in python3
result.sort(key=lambda x:x[0])

print(result)

输出

在python 2中

ls=[['01',2,'a',4],['02',5,2,4],['03',2,3,4],['01',1,'asd',4]]
totals = {}
for element in ls:
    key ,array = element[0],element[1:]
    for index in range(len(array)):
        if key not in totals:
            totals.setdefault(key,array)    
        elif (type(totals[key][index]) == "int") and type(array[index] == "int"):    
            totals[key][index] += array[index]    
result = []
for key,value in totals.items():
    result.append([key] + value)
print(result)

# you can sort the result to get exactly as you wanted
# you dont need this step if the dictionary is already sorted like in python3
result.sort(key=lambda x:x[0])

print(result)

在python 3中

[['02', 5, 2, 4], ['03', 2, 6, 4], ['01', 3, 6, 8]]
[['01', 3, 6, 8], ['02', 5, 2, 4], ['03', 2, 6, 4]]

Answer 3

如果我收到您的问题，则您实际上是在尝试按数组求和的方式进行分组。就个人而言，我会这样做：

sum_dictionary = {}

for ls in your_list:
    key = ls[0]
    vals = ls[1:]
    if key not in sum_dictionary:
        sum_dictionary[key] = [vals]
    else:
        sum_dictionary[key].append(vals)

for k, v in sum_dictionary.items():
    sum_dictionary[k] = list(map(sum, zip(*v)))

Answer 4

使用groupby from itertools也许更优雅一些：

steps {
    sh'''
      ...

      AA=XXX
      BB=YYY

      set > vars.prop
    '''

    script {
        vars = readProperties file: 'vars.prop'

        env << vars // merge vars into env

        echo 'AA='+ env['AA']
    }
}

请注意，上述解决方案将对原始列表from itertools import groupby from operator import itemgetter ls = [['01', 2, 3, 4], ['02', 5, 2, 4], ['03', 2, 6, 4], ['01', 1, 3, 4]] res = [] for k, group in groupby(sorted(ls), itemgetter(0)): group = [sub[1:] for sub in group] res.append([k] + [sum(x) for x in zip(*group)]) print(res) # -> [['01', 3, 6, 8], ['02', 5, 2, 4], ['03', 2, 6, 4]]进行排序！

子列表切片的有条件的，按术语的添加

4 个答案: