python to datetime_python-pandas to_datetime()未检测到列

加起来：

正如文档所说,重命名列的方法已经很聪明了：

Examples

Assembling a datetime from multiple columns of a DataFrame. The keys

can be common abbreviations like [‘year’, ‘month’, ‘day’, ‘minute’,

‘second’, ‘ms’, ‘us’, ‘ns’]) or plurals of the same

但是,还有其他选择.以我的经验,使用zip进行列表理解非常快(对于小集合).随着大约3000行数据的重命名,列变得最快.从图中可以看出,重命名的惩罚对于小集合来说很难,但对于大集合则可以弥补.

备择方案

pd.to_datetime(['-'.join(map(str,i)) for i in zip(df['h3'],df['h2'],df['h1'])])

pd.to_datetime(['-'.join(i) for i in df[['h3', 'h2', 'h1']].values.astype(str)])

df[['h3','h2','h1']].astype(str).apply(lambda x: pd.to_datetime('-'.join(x)), 1)

pd.to_datetime(df[['h1','h2','h3']].rename(columns={'h1':'day', 'h2':'month','h3':'year'}))

时间Win10：

#df = pd.concat([df]*1000)

2.74 ms ± 33.7 ?s per loop (mean ± std. dev. of 7 runs, 100 loops each)

8.08 ms ± 158 ?s per loop (mean ± std. dev. of 7 runs, 100 loops each)

158 ms ± 472 ?s per loop (mean ± std. dev. of 7 runs, 10 loops each)

2.64 ms ± 104 ?s per loop (mean ± std. dev. of 7 runs, 100 loops each)

MacBook Air的计时：

100 loops, best of 3: 6.1 ms per loop

100 loops, best of 3: 12.7 ms per loop

1 loop, best of 3: 335 ms per loop

100 loops, best of 3: 4.7 ms per loop

使用我编写的代码进行更新(如果您有改进的建议或任何可以帮助您的库,请感到高兴)：

import pandas as pd

import numpy as np

import timeit

import matplotlib.pyplot as plt

from collections import defaultdict

df = pd.DataFrame({

'h1': np.arange(1,11),

'h2': np.arange(1,11),

'h3': np.arange(2000,2010)

})

myfuncs = {

"pd.to_datetime(['-'.join(map(str,i)) for i in zip(df['h3'],df['h2'],df['h1'])])":

lambda: pd.to_datetime(['-'.join(map(str,i)) for i in zip(df['h3'],df['h2'],df['h1'])]),

"pd.to_datetime(['-'.join(i) for i in df[['h3','h2', 'h1']].values.astype(str)])":

lambda: pd.to_datetime(['-'.join(i) for i in df[['h3','h2', 'h1']].values.astype(str)]),

"pd.to_datetime(df[['h1','h2','h3']].rename(columns={'h1':'day','h2':'month','h3':'year'}))":

lambda: pd.to_datetime(df[['h1','h2','h3']].rename(columns={'h1':'day','h2':'month','h3':'year'}))

}

d = defaultdict(dict)

step = 10

cont = True

while cont:

lendf = len(df); print(lendf)

for k,v in mycodes.items():

iters = 1

t = 0

while t < 0.2:

ts = timeit.repeat(v, number=iters, repeat=3)

t = min(ts)

iters *= 10

d[k][lendf] = t/iters

if t > 2: cont = False

df = pd.concat([df]*step)

pd.DataFrame(d).plot().legend(loc='upper center', bbox_to_anchor=(0.5, -0.15))

plt.yscale('log'); plt.xscale('log'); plt.ylabel('seconds'); plt.xlabel('df rows')

plt.show()

返回值：