我有一些非常大的文件，看起来像这样:
05/31/2012,15:30:00.029,1306.25,1,E,0,,1306.25
05/31/2012,15:30:00.029,1306.25,8,E,0,,1306.25
我可以使用以下方法轻松读取它们：
pd.read_csv(gzip.open("myfile.gz"), header=None,names=
["date","time","price","size","type","zero","empty","last"], parse_dates=[[0,1]])
有没有一种有效的方法将这样的日期解析成pandas时间戳？如果没有，是否有编写cython函数的指南，可以将其传递给date_parser=？
我尝试编写自己的解析器函数，但仍然需要太长时间才能完成我正在处理的项目。
cimport numpy as np
import pandas as pd
import datetime
import numpy as np
def convert_date_cython(np.ndarray date_vec, np.ndarray time_vec):
cdef int i
cdef int N = len(date_vec)
cdef out_ar = np.empty(N, dtype=np.object)
date = None
for i in range(N):
if date is None or date_vec[i] != date_vec[i - 1]:
dt_ar = map(int, date_vec[i].split("/"))
date = datetime.date(dt_ar[2], dt_ar[0], dt_ar[1])
time_ar = map(int, time_vec[i].split(".")[0].split(":"))
time = datetime.time(time_ar[0], time_ar[1], time_ar[2])
out_ar[i] = pd.Timestamp(datetime.datetime.combine(date, time))
return out_ar
对Michael WS的解决方案进行了改进:
pandas.Timestamp最好在Cython代码外执行。atoi和处理本地C字符串比Python函数稍快。datetime-lib调用的数量从2个减少到1个（+1个偶尔的日期）。NB！此代码中的日期顺序为日/月/年。
总的来说，该代码似乎比原始的convert_date_cython快大约10倍。然而，如果在read_csv之后调用此代码，则在SSD硬盘上由于读取开销，总时间的差异仅为几个百分点。我猜在普通HDD上，差异会更小。
cimport numpy as np
import datetime
import numpy as np
import pandas as pd
from libc.stdlib cimport atoi, malloc, free
from libc.string cimport strcpy
### Modified code from Michael WS:
### https://dev59.com/Xm3Xa4cB1Zd3GeqPfHyL#15812787
def convert_date_fast(np.ndarray date_vec, np.ndarray time_vec):
cdef int i, d_year, d_month, d_day, t_hour, t_min, t_sec, t_ms
cdef int N = len(date_vec)
cdef np.ndarray out_ar = np.empty(N, dtype=np.object)
cdef bytes prev_date = <bytes> 'xx/xx/xxxx'
cdef char *date_str = <char *> malloc(20)
cdef char *time_str = <char *> malloc(20)
for i in range(N):
if date_vec[i] != prev_date:
prev_date = date_vec[i]
strcpy(date_str, prev_date) ### xx/xx/xxxx
date_str[2] = 0
date_str[5] = 0
d_year = atoi(date_str+6)
d_month = atoi(date_str+3)
d_day = atoi(date_str)
strcpy(time_str, time_vec[i]) ### xx:xx:xx:xxxxxx
time_str[2] = 0
time_str[5] = 0
time_str[8] = 0
t_hour = atoi(time_str)
t_min = atoi(time_str+3)
t_sec = atoi(time_str+6)
t_ms = atoi(time_str+9)
out_ar[i] = datetime.datetime(d_year, d_month, d_day, t_hour, t_min, t_sec, t_ms)
free(date_str)
free(time_str)
return pd.to_datetime(out_ar)
%H-%M-%S的时间字符串的数量为24*60*60=86400。如果您的数据集行数比这个数量大得多或者您的数据包含大量重复的时间戳，那么在解析过程中添加缓存可以显著提高速度。
对于没有Cython的人来说，以下是纯Python的替代方案：
import numpy as np
import pandas as pd
from datetime import datetime
def parse_datetime(dt_array, cache=None):
if cache is None:
cache = {}
date_time = np.empty(dt_array.shape[0], dtype=object)
for i, (d_str, t_str) in enumerate(dt_array):
try:
year, month, day = cache[d_str]
except KeyError:
year, month, day = [int(item) for item in d_str[:10].split('-')]
cache[d_str] = year, month, day
try:
hour, minute, sec = cache[t_str]
except KeyError:
hour, minute, sec = [int(item) for item in t_str.split(':')]
cache[t_str] = hour, minute, sec
date_time[i] = datetime(year, month, day, hour, minute, sec)
return pd.to_datetime(date_time)
def read_csv(filename, cache=None):
df = pd.read_csv(filename)
df['date_time'] = parse_datetime(df.loc[:, ['date', 'time']].values, cache=cache)
return df.set_index('date_time')
根据特定数据集，加速比为150x+：
$ ls -lh test.csv
-rw-r--r-- 1 blurrcat blurrcat 1.2M Apr 8 12:06 test.csv
$ head -n 4 data/test.csv
user_id,provider,date,time,steps
5480312b6684e015fc2b12bc,fitbit,2014-11-02 00:00:00,17:47:00,25
5480312b6684e015fc2b12bc,fitbit,2014-11-02 00:00:00,17:09:00,4
5480312b6684e015fc2b12bc,fitbit,2014-11-02 00:00:00,19:10:00,67
In [1]: %timeit pd.read_csv('test.csv', parse_dates=[['date', 'time']])
1 loops, best of 3: 10.3 s per loop
In [2]: %timeit read_csv('test.csv', cache={})
1 loops, best of 3: 62.6 ms per loop
为了限制内存使用，可以将字典缓存替换为类似于 LRU 的东西。
pd.Timestamp无法正常工作（但单独使用例如pd.Timestamp('05/31/2012,15:30:00.029')可以正常工作）。它不能正常工作的原因很可能是一个错误。 - Andy Hayden