只适用于列表列表中前两个列表的函数

Question

我有这个清单：

mylist = [
    [1890731350060, 'February 2016, March 2016, January 2016', 'INDEMNIZATIA DE HRANA', 1183], 
    [1890922350110, 'May 2015, June 2015, April 2015', 'INDEMNIZATIA DE HRANA', 1183], 
    [1890731350060, 'February 2016, March 2016, January 2016', 'INDEMNIZATIA DE HRANA', 1183]
]

我想要的结果是：

mylist = [
    [1890731350060, 'Ian 2016, Feb 2016, Mar 2016', 'INDEMNIZATIA DE HRANA', 1183],
    [1890922350110, 'Iun 2016, Mai 2016, Apr 2016', 'INDEMNIZATIA DE HRANA', 1183],
    [1890731350060, 'Ian 2016, Feb 2016, Mar 2016', 'INDEMNIZATIA DE HRANA', 1183]
]

为此，我有两个函数：

from datetime import datetime
import re
def translateInRo(string, dyct):
    substrs = sorted(dyct, key=len, reverse=True)

    regexp = re.compile('|'.join(map(re.escape, substrs)))

    return regexp.sub(lambda match: dyct[match.group(0)], string)

def orderDateslist(thislist):
    i=0
    for dates in thislist:
        sorted_list = []
        chgDates = dates[1].split(",")
        for test1 in chgDates:
            sorted_list.append(test1.strip())
        test = sorted(sorted_list, key=lambda x: datetime.strptime(x, "%B %Y"))
        str1 = ', '.join(test)
        translate = translateInRo(
            str1, {"January": "Ian", "February": "Feb", "March": "Mar", "April": "Apr", "May": "Mai", "June": "Iun", "July": "Iul", "August": "Aug", "September": "Sept", "October": "Oct", "November": "Nov", "December": "Dec"})
        thislist[i][1] = translate
        i = + 1
    return thislist

当我打印时：

print (orderDateslist(mylist))
[[1890731350060, 'Ian 2016, Feb 2016, Mar 2016', 'INDEMNIZATIA DE HRANA', 1183], [1890922350110, 'Ian 2016, Feb 2016, Mar 2016', 'INDEMNIZATIA DE HRANA', 1183], [1890731350060, 'February 2016, March 2016, January 2016', 'INDEMNIZATIA DE HRANA', 1183]]

最后一个列表不会被计算，我的函数只适用于列表列表中的前两个列表，之后的列表将保持不变，我希望这个函数适用于大量列表，我必须更改什么？我用的是python3。最后一个是复制。你知道吗

Answer 1

问题

为了澄清这个问题，从预期的代码来看，您似乎希望将每个子列表的索引1处的日期字符串替换为：

1. 按时间排序日期
2. 根据翻译词典缩写月份

可按以下步骤进行：

# Given 
import datetime


mylist = [
    [1890731350060, 'February 2016, March 2016, January 2016', 'INDEMNIZATIA DE HRANA', 1183], 
    [1890922350110, 'May 2015, June 2015, April 2015',         'INDEMNIZATIA DE HRANA', 1183], 
    [1890731350060, 'February 2016, March 2016, January 2016', 'INDEMNIZATIA DE HRANA', 1183]
]

TRANSLATE = {
    "January": "Ian", "February": "Feb", "March": "Mar", "April": "Apr",
    "May": "Mai", "June": "Iun", "July": "Iul", "August": "Aug", 
    "September": "Sept", "October": "Oct", "November": "Nov", "December": "Dec"
}

代码

def transform_dates(iterable, translate=TRANSLATE):
    transformed_lists = []
    for i, sublst in enumerate(iterable):
        transformed_lists.append(sublst[:])

        # Clean dates string
        raw_dates = sublst[1]
        cleaned_dates = set(map(str.strip, raw_dates.split(",")))

        # Sort dates string
        months_yrs = sorted(cleaned_dates, key=lambda x: datetime.datetime.strptime(x, "%B %Y"))
        months_yrs_split = [i.split() for i in months_yrs]

        # Abbreviate months
        abbrev_dates = [" ".join((translate[i[0]], i[1])) for i in months_yrs_split]
        transformed_lists[i][1] = ", ".join(abbrev_dates)
    return transformed_lists

transform_dates(mylist)
# [[1890731350060, 'Ian 2016, Feb 2016, Mar 2016', 'INDEMNIZATIA DE HRANA',1183],
#  [1890922350110, 'Apr 2015, Mai 2015, Iun 2015', 'INDEMNIZATIA DE HRANA',1183],
#  [1890731350060, 'Ian 2016, Feb 2016, Mar 2016', 'INDEMNIZATIA DE HRANA',1183]]

---

注意事项

此函数按月份和年份排序。你知道吗

lst = [1890731350060, 'February 2015, March 2013, January 2016', 'INDEMNIZATIA DE HRANA', 1183], 
transform_dates(lst)
# [[1890731350060, 'Mar 2013, Feb 2015, Ian 2016', 'INDEMNIZATIA DE HRANA', 1183]]

此函数用于删除重复的日期。你知道吗

lst = [1890731350060, 'May 2016, June 2016, May 2016, July 2016', 'INDEMNIZATIA DE HRANA', 1183], 
transform_dates(lst)
# [[1890731350060,'Mai 2016, Iun 2016, Iul 2016', 'INDEMNIZATIA DE HRANA', 1183]]

细节

如果您是Python新手，我将添加这些细节来帮助您表达正在发生的事情。你知道吗

transform_dates()函数接受名为mylist的列表列表作为and参数。在函数内部，我们首先创建一个名为transformed_lists的新列表，稍后将向其追加项。我们现在循环iterable（相当于mylist）以获得每个sublist，并跟踪它们的索引位置（i）。你知道吗

我们将sublst的一个副本添加到transform_dates（因此[:]，因为这样可以防止我们修改mylist中的原始项）。然后我们开始处理包含日期字符串的第一个索引。我们清理字符串，首先将其拆分为一个月-年对列表，然后strip尾随和前导空格，例如['February 2016', 'March 2016', 'January 2016']。如果有任何重复的日期，set()会删除它们，因为集合是唯一元素的集合。你知道吗

为了准备下一步，我们利用这个机会将它们按日期和split按单个空格进一步排序。拆分生成一个临时嵌套列表，例如[['January', '2016'], ['February', '2016'], ['March', '2016']]。你知道吗

最后，对于后一个嵌套列表中的每个项目，我们使用TRANSLATE字典缩写月份，并将其与年份一起join()，形成一个新字符串的单一列表，例如['Jan 2016', 'Feb 2016', 'Mar 2016']。然后我们执行最后一个join()，其中每个项由逗号分隔（根据请求），例如'Jan 2016, Feb 2016, Mar 2016'。你知道吗

我们已经完成了弦的变换。现在我们只需将新字符串赋给transformed_lists的索引1来替换旧字符串。总之，我们系统地选择了字符串，对其进行分解、转换、重新组合并将其重新分配到列表中的原始位置。我们对iterable中的每个sublist重复这个过程，直到循环完成。结果是我们的transformed_lists，它是由函数返回的。你知道吗

Answer 2

你可以试试这个：

import re
import itertools

def orderdates(full_date):
    table = {"January": "Ian", "February": "Feb", "March": "Mar", "April": "Apr", "May": "Mai", "June": "Iun", "July": "Iul", "August": "Aug", "September": "Sept", "October": "Oct", "November": "Nov", "December": "Dec"}
    l = ["Ian", "Feb", "Mar", "Apr", "Mai", "Iun", "Iul", "Aug", "Sept", "Oct", "Nov", "Dec"]
    new_dates = re.split(",\s", full_date)
    final_dates = [[a, int(b)] for a, b in [i.split() for i in new_dates]]

    new_dates = sorted(final_dates, key = lambda x: x[-1])

    current = [list(b) for a, b in itertools.groupby(new_dates, lambda x: x[-1])]
    new_current = [[table[i]+" "+str(b) for i, b in c] for c in current]
   final_current = [sorted(b, key= lambda x:l.index(x.split()[0])) for b in new_current]
  return list(itertools.chain.from_iterable(final_current))


mylist = [[1890731350060, 'January 2016, February 2016, March 2015', 'INDEMNIZATIA DE HRANA', 1183], [1890922350110, 'May 2015, June 2015, April 2015', 'INDEMNIZATIA DE HRANA', 1183], [1890731350060, 'February 2016, March 2016, January 2016', 'INDEMNIZATIA DE HRANA', 1183]]

new_data = [[i[0], orderdates(i[1]), i[2:]] for i in mylist]

new_data = [list(itertools.chain(*[[b] if not isinstance(b, list) else b for b in i])) for i in new_data]
print(new_data)

输出：

[[1890731350060, 'Mar 2015', 'Ian 2016', 'Feb 2016', 'INDEMNIZATIA DE HRANA', 1183], [1890922350110, 'Apr 2015', 'Mai 2015', 'Iun 2015', 'INDEMNIZATIA DE HRANA', 1183], [1890731350060, 'Ian 2016', 'Feb 2016', 'Mar 2016', 'INDEMNIZATIA DE HRANA', 1183]]