Question

我想知道如何最好地在数据帧中有条件地拆分字符串，以便正确的值到达正确的列。当字符串的长度变化时，请考虑以下内容。时间以小时，分钟，秒为单位。

time = ['22:35', '1:20:20']
df = pd.DataFrame({'time': time})
df
    time
0   22:35
1   1:20:20

对此进行了尝试：

for i in df['time']:
    if len(i) == 5:
        df[['minutes', 'seconds']] = df['time'].str.split(':', expand=True)
    if len(i) == 7:
        df[['hours', 'minutes', 'seconds']] = df['time'].str.split(':', expand=True)

Answer 1

可能不是一个很好的答案，但我相信这是可行的：

time = ['22:35', '1:20:20']

templist = [item.split(":") for item in time]

templist2 = []
for item in templist:
    if len(item) == 3:
        thisdict = {
            "hours": item[0],
            "minutes": item[1],
            "seconds": item[2]
        }
        templist2.append(thisdict)
    elif len(item) == 2:
        thisdict = {
            "hours": "0",
            "minutes": item[0],
            "seconds": item[1]
        }
        templist2.append(thisdict)
    elif len(item) == 1:
        thisdict = {
            "hours": "0",
            "minutes": "0",
            "seconds": item[0]
        }
        templist2.append(thisdict)

df = pd.DataFrame(templist2)

Answer 2

我同意@Mark Wang和@ s.k，但这可能对您有用：

import datetime

def time_converter(s):
    nb_semicolons = s.count(':')
    if nb_semicolons == 1:
        form = '%M:%S'
    if nb_semicolons == 2:
        form = '%H:%M:%S'
    return datetime.datetime.strptime(s, form).time()

df['time2'] = df['time'].apply(time_converter)

Answer 3

事实证明这很容易：

import pandas as pd
time = ['22:35', '1:20:20']
df = pd.DataFrame({'time': time})

def f1(x):
    if len(x) == 5:
        x = '0:' + x
    return x

df['time'] = df['time'].apply(f1)
df[['hours', 'minutes', 'seconds']] = df['time'].str.split(':', expand=True)
df

    time    hours   minutes seconds
0   0:22:35 0         22     35
1   1:20:20 1         20     20

熊猫：有条件str.split

3 个答案: