我有三列(h1、h2、h3)分别代表日、月和年,例如
import pandas as pd
df = pd.DataFrame({
'h1': [1,2,3],
'h2': [1,2,3],
'h3': [2000,2001,2002]
})
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当我执行:
pd.to_datetime(df[['h1', 'h2', 'h3']])
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这导致错误:ValueError: to assemble mappings requires at least that [year, month, day] be specified: [day,month,year] is missing 但是当我重命名列然后执行 pd.to_datetime 例如
df=df.rename(columns ={'h1':'day', 'h2':'month', 'h3': 'year'})
df["date_col"] =pd.to_datetime(df[['day','month','year']])
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在上面我得到了年份栏,我们必须这样做吗?或者是否可以提供一种格式,以便可以分别将列检测为日、月、年?谢谢。
正如文档所说,您重命名列的方法已经很聪明:
例子
从 DataFrame 的多列组装日期时间。键可以是常见的缩写,如 ['year', 'month', 'day', 'minute', 'second', 'ms', 'us', 'ns']) 或相同的复数
但是有一些替代方案。根据我的经验,使用 zip 的列表理解非常快(对于小集合)。使用大约 3000 行数据重命名列成为最快的。查看图表,重命名的惩罚对于小集合很难,但可以补偿大集合。
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'}))
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#df = pd.concat([df]*1000)
2.74 ms ± 33.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
8.08 ms ± 158 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
158 ms ± 472 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
2.64 ms ± 104 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
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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
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更新我写的代码(如果你有改进建议或任何可以帮助的库,我很高兴):
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()
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