format()：ValueError：整数格式说明符中不允许使用精度。

Question

我是一名Python新手。我正在熟悉格式化方法。

这是我正在阅读的一本学习Python的书。

What Python does in the format method is that it substitutes each argument
value into the place of the specification. There can be more detailed specifications
such as:
decimal (.) precision of 3 for float '0.333'
>>> '{0:.3}'.format(1/3)
fill with underscores (_) with the text centered
(^) to 11 width '___hello___'
>>> '{0:_^11}'.format('hello')
keyword-based 'Swaroop wrote A Byte of Python'
>>> '{name} wrote {book}'.format(name='Swaroop', book='A Byte of Python')

如果我在Python解释器中尝试执行以下命令：

print('{0:.3}'.format(1/3))

它会报错

 File "", line 24, in 
ValueError: Precision not allowed in integer format specifier 

- liv2hak

Answer 1

最好添加f：

In [9]: print('{0:.3f}'.format(1/3))
0.000

通过这种方式，您可以注意到1/3产生了一个整数，然后将其更正为1./3或1/3.。

- zhangxaochen

Answer 2

要打印浮点数，您必须将至少一个输入设置为浮点数，就像这样

print('{0:.3}'.format(1.0/3))

如果除数和被除数均为整数，则返回的结果也将是整数，小数部分将被舍去。

输出

0.333

你可以使用float函数将数据转换为浮点数，像这样。

data = 1
print('{0:.3}'.format(float(data) / 3))

- thefourtheye

---

"{0:.3}" 是什么意思？格式化如何替换这些值？ - liv2hak

@liv2hak，在格式函数中，0是第一个参数。冒号只是传递数据元素和格式化之间的分隔符。.3表示将其格式化为小数点右侧3个字符。 - John

Answer 3

值得注意的是，这个错误只会在Python 2中发生。在Python 3中，除法总是返回一个浮点数。

您可以通过在Python 2中使用 "from __future__ import division" 语句来复制此错误。

~$ python
Python 2.7.6 
>>> '{0:.3}'.format(1/3)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: Precision not allowed in integer format specifier
>>> from __future__ import division
>>> '{0:.3}'.format(1/3)
'0.333'

- Jacklynn

Answer 4

Python3 fstring示例：

小数

# Set the Result to a variable:
a = 1/3
# Format the result :.3f is the length of the digits "AFTER" the decial point.
print(f"{a:.3f}")
# Returns: 0.333

print(f"{a:.6f}")
# Returns: 0.333333

print(f"{a:.32f}")
# Returns: 0.33333333333333331482961625624739

print(f"{a:.50f}")
# Returns: 0.33333333333333331482961625624739099293947219848633

print(f"{a:.55f}")
# Returns: 0.3333333333333333148296162562473909929394721984863281250

# # NOTE: this does round see the difference between the ending of .50 & .55

数字和小数

# You can do the same for leading zeros:
a = (1/3) + 145630

print(f"{a:016.03f}")

# Returns 000000145630.333

# if you don't want to lead with Zeros and just spaces.
print(f"{a:16.03f}")
# Returns "      145630.333"
# # NOTE: 
# With this one - notice there are only 12 Digits/Spaces Left of the decimal.

print(f"{a:016.55f}")
# Returns 145630.3333333333430346101522445678710937500000000000000000000
# # NOTE:
# will never show leading spaces or zeros
# as the decimal digit count is greater than the total digit count.

# So the way to calculate it what will properly be displayed is:
# `f"{VARIABLE:(TOTALDIGITS).(DECIMALDIGITS)f}"`
# Total Digits - Decimal Digits - 1 (for the decimal point)
# If the return from the equasion above is < 1 
# # There will never be a leading digit/space.  
# As shown in the last example of Digits & Decimals.  (16-55-1) =
# "-40"  So no Leading digits will ever show.

仅限数字

# If you only need to update formats for Digits and no Decimals:
a = 148
print(f"{a:016d}")
# or
print(f"{a:016.0f}")

# Returns 0000000000000148

print(f"{a:16d}")
# or 
print(f"{a:16.0f}")
# Returns "             148"

需要注意的异常情况：

a = 148.15
print(f"{a:16d}")  # THROWS AN ERROR:
# ValueError: Unknown format code 'd' for object of type 'float'

# or 
print(f"{a:16.0f}")
# Returns "             148"
# # NOTE: This TRUNCATES YOUR DECIMAL FORMAT.

- JayRizzo