Question

我目前正在使用以下内容来计算两次差异。外出非常快，因此无论如何我都不需要显示仅为0.00的小时和分钟。我如何在Python中实际移位小数位？

def time_deltas(infile): 
    entries = (line.split() for line in open(INFILE, "r")) 
    ts = {}  
    for e in entries: 
        if " ".join(e[2:5]) == "OuchMsg out: [O]": 
            ts[e[8]] = e[0]    
        elif " ".join(e[2:5]) == "OuchMsg in: [A]":    
            in_ts, ref_id = e[0], e[7] 
            out_ts = ts.pop(ref_id, None) 
            yield (float(out_ts),ref_id[1:-1], "%.10f"%(float(in_ts) - float(out_ts)))

INFILE = 'C:/Users/kdalton/Documents/Minifile.txt'
print list(time_deltas(INFILE))

Answer 1

与数学相同

a = 0.01;
a *= 10; // shifts decimal place right
a /= 10.; // shifts decimal place left

Answer 2

或使用datetime模块

>>> import datetime
>>> a = datetime.datetime.strptime("30 Nov 11 0.00.00", "%d %b %y %H.%M.%S")
>>> b = datetime.datetime.strptime("2 Dec 11 0.00.00", "%d %b %y %H.%M.%S")
>>> a - b
datetime.timedelta(-2)

Answer 3

def move_point(number, shift, base=10):
    """
    >>> move_point(1,2)
    100
    >>> move_point(1,-2)
    0.01
    >>> move_point(1,2,2)
    4
    >>> move_point(1,-2,2)
    0.25
    """
    return number * base**shift

Answer 4

扩大接受的答案；这是一个函数，它将给您提供的任何数字移动到小数位。如果小数位参数为0，则返回原始数字。为了一致性，总是返回float类型。

def moveDecimalPoint(num, decimal_places):
    '''
    Move the decimal place in a given number.

    args:
        num (int)(float) = The number in which you are modifying.
        decimal_places (int) = The number of decimal places to move.
    
    returns:
        (float)
    
    ex. moveDecimalPoint(11.05, -2) returns: 0.1105
    '''
    for _ in range(abs(decimal_places)):

        if decimal_places>0:
            num *= 10; #shifts decimal place right
        else:
            num /= 10.; #shifts decimal place left

    return float(num)

print (moveDecimalPoint(11.05, -2))

如何在Python中移动小数位？

4 个答案: