如何以良好的格式结果格式化 python 列表

我有很多列表，如下所示，通过打印data并将其与字符串连接起来check_str如下最后一条语句所示。需要帮助以在字符串连接后更好地格式化列表值，如预期所示

print(data) #printing lists in this variable

['0 rows returned']
[(Decimal('1456'),datetime.date(2022, 9, 15)), (Decimal('156'),datetime.date(2012, 10, 15))]
['0 rows returned']
['0 rows returned']
[(Decimal('1256'),datetime.date(2112, 9, 15)), (Decimal('56'),datetime.date(2012, 10, 25)), (Decimal('561'),datetime.date(2012, 10, 25))]

用字符串连接每个列表check_str

check_str = 'data is there'
#Below is the last statement where I am concatenating string with each list from data variable
print(check_str + '|' + ','.join(str(v) for v in data)) 

上述打印语句的实际输出：

data is there|0 rows returned
data is there|(Decimal('1456'),datetime.date(2022, 9, 15))(Decimal('156'),datetime.date(2012, 10, 15))
data is there|0 rows returned
data is there|0 rows returned
data is there|(Decimal('1256'),datetime.date(2112, 9, 15)), (Decimal('56'),datetime.date(2012, 10, 25)), (Decimal('561'),datetime.date(2012, 10, 25))

预期：如果可能，需要采用以下良好格式的上述输出。

data is there|0 rows returned
data is there|1456|2022-09-15
             |156|2022-10-15
data is there|0 rows returned
data is there|0 rows returned
data is there|1256|2112-09-15
             |56|2012-10-25
             |561|2012-10-25

你的问题是你正在申请str()在元组上，而不是在小数和日期值上。

您需要解压元组才能获得所需的输出。

这是一种解决方案：

from decimal import *
import datetime

list_of_lists = [
    ['0 rows returned'],
    [
        (Decimal('1256'),datetime.date(2112, 9, 15)), 
        (Decimal('56'),datetime.date(2012, 10, 25)), 
        (Decimal('561'),datetime.date(2012, 10, 25))
    ]
]

check_str = 'data is there'
pad_str =   '             '

for l in list_of_lists:

    if l == ['0 rows returned']:
        print('%s|%s'%(check_str, l[0]))
    else:
        prepend = check_str
        for t in l:
             print('%s|%s'%(prepend,'|'.join(str(v) for v in t)))
             prepend = pad_str