如何计算几个数字的位数，然后将它们格式化为表格形式？
How do I count the number of digits of several numbers and then format them in tabular form?
我正在为 class 编写一个非常简单的程序，它根据固定票价计算收入。乘以用户提供的票数后，先在千位用逗号格式化，然后在前面加美元符号，最后returns为要列在的字符串顶部有固定标题的列。
我想知道有没有办法先统计最后生成的字符串的位数（包括美元符号和逗号），然后据此调整列的宽度，如果空格数比正确显示列的 header 的最小空格数长，如果这有意义的话。
仅供参考，脚本已经输出了项目所需的结果，我只是想进一步扩展我自己的理解。另外，我知道这可能不是构建此程序的最佳方式，因此欢迎任何建议：
# Define ticket prices
SILVER_TIC_PRICE = 35
GOLD_TIC_PRICE = 55
PLATINUM_TIC_PRICE = 85
# Create Functions
# Validate tickets sold as an interger
def tickets_sold(message):
while True:
try:
user_input = int(input(message))
except ValueError:
print('\nERROR : \nInvalid entry \nPlease use a number\n')
continue
try:
assert user_input >= 0
except AssertionError:
print('\nERROR : \nInvalid entry \nPlease use a positive number\n')
continue
else:
return user_input
break
# Reusable line function
def print_line(ticket_class, tickets_sold, ticket_revenue):
print(format(ticket_class, " >8"), format(tickets_sold, " >7"), format(ticket_revenue, " <7"), sep=' ')
# Format Function
# def format_final()
# Get the number of tickets sold by type
# But first a blank line for readability
print(' ')
# Get Silver tickets sold
silver_tickets_sold = tickets_sold ('Enter the number of Silver tickets sold: ')
# Get Gold tickets sold
gold_ticket_sold = tickets_sold('Enter the number of Gold tickets sold: ')
# Get Platinum tickets sold
platinum_ticket_sold = tickets_sold('Enter the number of Platinum tickets sold: ')
# calculate revenue
silver_initial = int(silver_tickets_sold * SILVER_TIC_PRICE)
gold_initial = int(gold_ticket_sold * GOLD_TIC_PRICE)
platinum_initial = int(platinum_ticket_sold * PLATINUM_TIC_PRICE)
silver_final = "{:,}".format(silver_initial)
gold_final = "{:,}".format(gold_initial)
platinum_final = "{:,}".format(platinum_initial)
silver_revenue = '$' + str(silver_final)
gold_revenue = '$' + str(gold_final)
platinum_revenue = '$' + str(platinum_final)
# calculate totals
total_tickets = int(silver_tickets_sold + gold_ticket_sold + platinum_ticket_sold)
total_initial = int(silver_initial + gold_initial + platinum_initial)
total_final = "{:,}".format(total_initial)
total_revenue = '$'+str(total_final)
# display results
print(' ')
print_line('Section','Tickets','Revenue')
print_line('--------','-------','-------')
print_line('Silver', silver_tickets_sold, silver_revenue)
print_line('Gold', gold_ticket_sold,gold_revenue)
print_line('Platinum', platinum_ticket_sold, platinum_revenue)
print_line('========','=======','=======')
print_line('Total', total_tickets, total_revenue)
print(' ')
我知道已经有一些问题可以回答其中的部分问题，但我找不到任何能够以我能够全神贯注的方式组合这些部分的内容。
如果这个数字你只对大小感兴趣，那么它是以十为底的。
import math
math.ceil(math.log(n,10))
应该可以。
或者您可以使用以下任一方法：
len(str(n))
len(str(n).replace(".",""))
我整理了一个小代码，可以计算出最终字符串中数字的个数 total_revenue。
# Initialize the counter
count_number = 0
# Get the count of numbers
for c in total_revenue:
if(c.isdigit()):
count_number += 1
# Display the count
print(count_number)
好吧，让我从一些代码审查开始：
第一个功能很好。它正确使用 while True 和 break 直到给出有效数字。但是还有一些改进的余地：
def tickets_sold(message):
while True:
try:
user_input = int(input(message))
except ValueError:
print('\nERROR : \nInvalid entry \nPlease use a number\n')
continue
if user_input < 0: # better to use an if instead of "try: assert"
print('\nERROR : \nInvalid entry \nPlease use a positive number\n')
continue
return user_input # just return all other cases already "continue"d
然而，它从大量重复开始，我会输入请求门票的代码，然后 returns 门票数量和收入。通过将其中一些放入函数中，可以简单地避免很多（不是全部）重复：
def get_tickets_and_revenue(name, tic_price):
n_tickets_sold = tickets_sold('Enter the number of {} tickets sold: '.format(name))
revenue = int(n_tickets_sold * tic_price)
return n_tickets_sold, revenue
silver_tickets_sold, silver_revenue = get_tickets_and_revenue('Silver', SILVER_TIC_PRICE)
gold_ticket_sold, gold_revenue = get_tickets_and_revenue('Gold', GOLD_TIC_PRICE)
platinum_ticket_sold, platinum_revenue = get_tickets_and_revenue('Platinum', PLATINUM_TIC_PRICE)
total_tickets = silver_tickets_sold + gold_ticket_sold + platinum_ticket_sold
total_revenue = silver_revenue + gold_revenue + platinum_revenue
def format_number_according_to_spec(num):
return '${:,}'.format(num)
# only now convert them to strings!
silver_revenue = format_number_according_to_spec(silver_revenue)
gold_revenue = format_number_according_to_spec(gold_revenue)
platinum_revenue = format_number_according_to_spec(platinum_revenue)
total_revenue = format_number_according_to_spec(total_revenue)
下一部分可以通过将值保留为可迭代来缩短，但它也适用于硬编码：
def find_column_length(*column_values):
# using "len(str(sth))" you can determine the length!
return max(len(str(value)) for value in column_values)
# The row formatter is formatted twice, once with the length values and once
# with the actual values. So we need to escape the outer {} by doubling them.
row_formatter = '{{:>{}}} {{:>{}}} {{:>{}}}'
column_lengths = [find_column_length('Section',
'Silver',
'Gold',
'Platinum',
'Total'),
find_column_length('Tickets',
silver_tickets_sold,
gold_ticket_sold,
platinum_ticket_sold,
total_tickets),
find_column_length('Revenue',
silver_revenue,
gold_revenue,
platinum_revenue,
total_revenue)]
row_formatter = row_formatter.format(*column_lengths)
placeholder1 = '{} {} {}'.format('-'*column_lengths[0],
'-'*column_lengths[1],
'-'*column_lengths[2])
placeholder2 = '{} {} {}'.format('='*column_lengths[0],
'='*column_lengths[1],
'='*column_lengths[2])
print(' ')
print(row_formatter.format('Section','Tickets','Revenue'))
print(placeholder1)
print(row_formatter.format('Silver', silver_tickets_sold, silver_revenue))
print(row_formatter.format('Gold', gold_ticket_sold,gold_revenue))
print(row_formatter.format('Platinum', platinum_ticket_sold, platinum_revenue))
print(placeholder2)
print(row_formatter.format('Total', total_tickets, total_revenue))
print(' ')
例如：
Enter the number of Silver tickets sold: 200000000
Enter the number of Gold tickets sold: 1
Enter the number of Platinum tickets sold: 20
Section Tickets Revenue
-------- --------- --------------
Silver 200000000 ,000,000,000
Gold 1
Platinum 20 ,700
======== ========= ==============
Total 200000021 ,000,001,755
和
Enter the number of Silver tickets sold: 2
Enter the number of Gold tickets sold: 5
Enter the number of Platinum tickets sold: 1
Section Tickets Revenue
-------- ------- -------
Silver 2
Gold 5 5
Platinum 1
======== ======= =======
Total 8 0
我正在为 class 编写一个非常简单的程序，它根据固定票价计算收入。乘以用户提供的票数后，先在千位用逗号格式化，然后在前面加美元符号，最后returns为要列在的字符串顶部有固定标题的列。
我想知道有没有办法先统计最后生成的字符串的位数（包括美元符号和逗号），然后据此调整列的宽度，如果空格数比正确显示列的 header 的最小空格数长，如果这有意义的话。
仅供参考，脚本已经输出了项目所需的结果，我只是想进一步扩展我自己的理解。另外，我知道这可能不是构建此程序的最佳方式，因此欢迎任何建议：
# Define ticket prices
SILVER_TIC_PRICE = 35
GOLD_TIC_PRICE = 55
PLATINUM_TIC_PRICE = 85
# Create Functions
# Validate tickets sold as an interger
def tickets_sold(message):
while True:
try:
user_input = int(input(message))
except ValueError:
print('\nERROR : \nInvalid entry \nPlease use a number\n')
continue
try:
assert user_input >= 0
except AssertionError:
print('\nERROR : \nInvalid entry \nPlease use a positive number\n')
continue
else:
return user_input
break
# Reusable line function
def print_line(ticket_class, tickets_sold, ticket_revenue):
print(format(ticket_class, " >8"), format(tickets_sold, " >7"), format(ticket_revenue, " <7"), sep=' ')
# Format Function
# def format_final()
# Get the number of tickets sold by type
# But first a blank line for readability
print(' ')
# Get Silver tickets sold
silver_tickets_sold = tickets_sold ('Enter the number of Silver tickets sold: ')
# Get Gold tickets sold
gold_ticket_sold = tickets_sold('Enter the number of Gold tickets sold: ')
# Get Platinum tickets sold
platinum_ticket_sold = tickets_sold('Enter the number of Platinum tickets sold: ')
# calculate revenue
silver_initial = int(silver_tickets_sold * SILVER_TIC_PRICE)
gold_initial = int(gold_ticket_sold * GOLD_TIC_PRICE)
platinum_initial = int(platinum_ticket_sold * PLATINUM_TIC_PRICE)
silver_final = "{:,}".format(silver_initial)
gold_final = "{:,}".format(gold_initial)
platinum_final = "{:,}".format(platinum_initial)
silver_revenue = '$' + str(silver_final)
gold_revenue = '$' + str(gold_final)
platinum_revenue = '$' + str(platinum_final)
# calculate totals
total_tickets = int(silver_tickets_sold + gold_ticket_sold + platinum_ticket_sold)
total_initial = int(silver_initial + gold_initial + platinum_initial)
total_final = "{:,}".format(total_initial)
total_revenue = '$'+str(total_final)
# display results
print(' ')
print_line('Section','Tickets','Revenue')
print_line('--------','-------','-------')
print_line('Silver', silver_tickets_sold, silver_revenue)
print_line('Gold', gold_ticket_sold,gold_revenue)
print_line('Platinum', platinum_ticket_sold, platinum_revenue)
print_line('========','=======','=======')
print_line('Total', total_tickets, total_revenue)
print(' ')
我知道已经有一些问题可以回答其中的部分问题，但我找不到任何能够以我能够全神贯注的方式组合这些部分的内容。
如果这个数字你只对大小感兴趣，那么它是以十为底的。
import math
math.ceil(math.log(n,10))
应该可以。 或者您可以使用以下任一方法：
len(str(n))
len(str(n).replace(".",""))
我整理了一个小代码，可以计算出最终字符串中数字的个数 total_revenue。
# Initialize the counter
count_number = 0
# Get the count of numbers
for c in total_revenue:
if(c.isdigit()):
count_number += 1
# Display the count
print(count_number)
好吧，让我从一些代码审查开始：
第一个功能很好。它正确使用 while True 和 break 直到给出有效数字。但是还有一些改进的余地：
def tickets_sold(message):
while True:
try:
user_input = int(input(message))
except ValueError:
print('\nERROR : \nInvalid entry \nPlease use a number\n')
continue
if user_input < 0: # better to use an if instead of "try: assert"
print('\nERROR : \nInvalid entry \nPlease use a positive number\n')
continue
return user_input # just return all other cases already "continue"d
然而，它从大量重复开始，我会输入请求门票的代码，然后 returns 门票数量和收入。通过将其中一些放入函数中，可以简单地避免很多（不是全部）重复：
def get_tickets_and_revenue(name, tic_price):
n_tickets_sold = tickets_sold('Enter the number of {} tickets sold: '.format(name))
revenue = int(n_tickets_sold * tic_price)
return n_tickets_sold, revenue
silver_tickets_sold, silver_revenue = get_tickets_and_revenue('Silver', SILVER_TIC_PRICE)
gold_ticket_sold, gold_revenue = get_tickets_and_revenue('Gold', GOLD_TIC_PRICE)
platinum_ticket_sold, platinum_revenue = get_tickets_and_revenue('Platinum', PLATINUM_TIC_PRICE)
total_tickets = silver_tickets_sold + gold_ticket_sold + platinum_ticket_sold
total_revenue = silver_revenue + gold_revenue + platinum_revenue
def format_number_according_to_spec(num):
return '${:,}'.format(num)
# only now convert them to strings!
silver_revenue = format_number_according_to_spec(silver_revenue)
gold_revenue = format_number_according_to_spec(gold_revenue)
platinum_revenue = format_number_according_to_spec(platinum_revenue)
total_revenue = format_number_according_to_spec(total_revenue)
下一部分可以通过将值保留为可迭代来缩短，但它也适用于硬编码：
def find_column_length(*column_values):
# using "len(str(sth))" you can determine the length!
return max(len(str(value)) for value in column_values)
# The row formatter is formatted twice, once with the length values and once
# with the actual values. So we need to escape the outer {} by doubling them.
row_formatter = '{{:>{}}} {{:>{}}} {{:>{}}}'
column_lengths = [find_column_length('Section',
'Silver',
'Gold',
'Platinum',
'Total'),
find_column_length('Tickets',
silver_tickets_sold,
gold_ticket_sold,
platinum_ticket_sold,
total_tickets),
find_column_length('Revenue',
silver_revenue,
gold_revenue,
platinum_revenue,
total_revenue)]
row_formatter = row_formatter.format(*column_lengths)
placeholder1 = '{} {} {}'.format('-'*column_lengths[0],
'-'*column_lengths[1],
'-'*column_lengths[2])
placeholder2 = '{} {} {}'.format('='*column_lengths[0],
'='*column_lengths[1],
'='*column_lengths[2])
print(' ')
print(row_formatter.format('Section','Tickets','Revenue'))
print(placeholder1)
print(row_formatter.format('Silver', silver_tickets_sold, silver_revenue))
print(row_formatter.format('Gold', gold_ticket_sold,gold_revenue))
print(row_formatter.format('Platinum', platinum_ticket_sold, platinum_revenue))
print(placeholder2)
print(row_formatter.format('Total', total_tickets, total_revenue))
print(' ')
例如：
Enter the number of Silver tickets sold: 200000000
Enter the number of Gold tickets sold: 1
Enter the number of Platinum tickets sold: 20
Section Tickets Revenue
-------- --------- --------------
Silver 200000000 ,000,000,000
Gold 1
Platinum 20 ,700
======== ========= ==============
Total 200000021 ,000,001,755
和
Enter the number of Silver tickets sold: 2
Enter the number of Gold tickets sold: 5
Enter the number of Platinum tickets sold: 1
Section Tickets Revenue
-------- ------- -------
Silver 2
Gold 5 5
Platinum 1
======== ======= =======
Total 8 0