Python：文字列内の数値にのみ数値を加算/減算する方法は？

Question

言ってやる例えば、私は次の文字列と秒を表し入力4.0、持っている：

John Time Made 11:05:20 in 2010 
5.001 Kelly #1 
6.005 Josh #8 

をそして次の結果たいと思います：私は最初にhours:minutes:secondsを認識することができますどのように

John Time Made 11:05:24 in 2010 #Input 4.0 is added to the seconds of 11:05:20 
1.001 Kelly #1 #4.0 is subtracted from the first number 5.001 = 1.001 
2.005 Josh #8 #4.0 is subtracted from the first number 5.001 = 2.005 

を入力番号を追加/減算するには、残りの部分に#.###を追加しますか？

は

Answer 1

あなたの完全なデータは、あなたが提供されるこの特定のサンプルと同じフォーマットを持っている場合は、このソリューションが動作するはず答えupvote /事前にありがとうと受け入れます。 input.txtファイルにデータを格納する必要があります。 regexはコメントで落胆したが

val_to_add = 4 

with open('input.txt') as fin: 
    # processing first line 
    first_line = fin.readline().strip() 
    splitted = first_line.split(' ') 

    # get hour, minute, second corresponding to time (11:05:20) 
    time_values = splitted[3].split(':') 

    # seconds is the last element 
    seconds = int(time_values[-1]) 

    # add the value 
    new_seconds = seconds + val_to_add 

    # doing simple math to avoid having values >= 60 for minute and second 
    # this part probably can be solved with datetime or some other lib, but it's not that complex, so I did it in couple of lines 
    seconds = new_seconds % 60 # if we get > 59 seconds we only put the modulo as second and the other part goes to minute 
    new_minutes = int(time_values[1]) + new_seconds // 60 # if we have more than 60 s then here we'll add minutes produced by adding to the seconds 
    minutes = new_minutes % 60 # similarly as for seconds 
    hours = int(time_values[0]) + new_minutes // 60 

    # here I convert again to string so we could easily apply join operation (operates only on strings) and additionaly add zero in front for 1 digit numbers 
    time_values[0] = str(hours).rjust(2, '0') 
    time_values[1] = str(minutes).rjust(2, '0') 
    time_values[2] = str(seconds).rjust(2, '0') 

    new_time_val = ':'.join(time_values)# join the values to follow the HH:MM:SS format 
    splitted[3] = new_time_val# replace the old time with the new one (with the value added) 
    first_line_modified = ' '.join(splitted)# just join the modified list 

    print(first_line_modified) 

    # processing othe lines 
    for line in fin: 
     # here we only get the first (0th) value and subtract the val_to_add and round to 3 digits the response (to avoid too many decimal places) 
     stripped = line.strip() 
     splitted = stripped.split(' ') 
     splitted[0] = str(round(float(splitted[0]) - val_to_add, 3)) 
     modified_line = ' '.join(splitted) 
     print(modified_line) 

Answer 2

、regexは、必要な形式でそれらを印刷し、それらに必要な計算を実行し、datetime.timeオブジェクトに時刻オブジェクトを解析するために使用することができる。

# datetime module for time calculations 
import datetime 
# regex module 
import re 

# seconds to add to time 
myinp = 4 

# List of data strings 
# data = 'John Time Made 11:05:20 in 2010', '5.001 Kelly', '6.005 Josh' 

with open('data.txt') as f: 
    data = f.readlines() 

new_data = [] 

#iterate through the list of data strings 
for time in data: 
    try: 
     # First check for 'HH:MM:SS' time format in data string 
     # regex taken from this question: http://stackoverflow.com/questions/8318236/regex-pattern-for-hhmmss-time-string 
     match = re.findall("([0-1]?\d|2[0-3]):([0-5]?\d):([0-5]?\d)", time) 
     # this regex returns a list of tuples as strings "[('HH', 'MM', 'SS')]", 
     # which we join back together with ':' (colon) separators 
     t = ':'.join(match[0]) 
     # create a Datetime object from indexing the first matched time in the list, 
     # taken from this answer http://stackoverflow.com/questions/100210/what-is-the-standard-way-to-add-n-seconds-to-datetime-time-in-python 
     # May create an IndexError exception, which we catch in the `except` clause below 
     orig = datetime.datetime(100,1,1,int(match[0][0]), int(match[0][1]), int(match[0][2])) 
     # Add the number of seconds to the Datetime object, 
     # taken from this answer: http://stackoverflow.com/questions/656297/python-time-timedelta-equivalent 
     newtime = (orig + datetime.timedelta(0, myinp)).time() 
     # replace the time in the original data string with the newtime and print 
     new_data.append(time.replace(t, str(newtime))) 
    # catch an IndexError Exception, which we look for float-formatted seconds only 
    except IndexError: 
     # look for float-formatted seconds (s.xxx)  
     # taken from this answer: http://stackoverflow.com/questions/4703390/how-to-extract-a-floating-number-from-a-string 
     match = re.findall("\d+\.\d+", time) 
     # create a Datetime object from indexing the first matched time in the list, 
     # specifying only seconds, and microseconds, which we convert to milliseconds (micro*1000) 
     orig = datetime.datetime(100,1,1,second=int(match[0].split('.')[0]),microsecond=int(match[0].split('.')[1])*1000) 
     # Subtract the seconds from the Datetime object, similiar to the time addtion in the `try` clause above 
     newtime = orig - datetime.timedelta(0, myinp) 
     # format the newtime as `seconds` concatenated with the milliseconds converted from microseconds 
     newtime_fmt = newtime.second + newtime.microsecond/1000000. 
     # Get the seconds value (first value(index 0)) from splitting the original string at the `space` between the `seconds` and `name` strings 
     t = time.split(' ')[0] 
     # replace the time in the original data string with the newtime and print 
     new_data.append(time.replace(t , str(newtime_fmt))) 


with open('new_data.txt', 'w') as nf: 
    for newline in new_data: 
     nf.write(newline) 

new_data.txtファイルの内容は次のようになります。

John Time Made 11:05:24 in 2010 
1.001 Kelly 
2.005 Josh