Python 在熊猫中将浮点系列转换为整数系列
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Convert float Series into an integer Series in pandas
提问by Geekster
I have the following data frame:
我有以下数据框：
In [31]: rise_p
Out[31]:
time magnitude
0 1379945444 156.627598
1 1379945447 1474.648726
2 1379945448 1477.448999
3 1379945449 1474.886202
4 1379945699 1371.454224
Now, I want to group rows which are within a minute. So I divide the time series with 100. I get this:
现在，我想对一分钟内的行进行分组。所以我将时间序列除以 100。我得到了这个：
In [32]: rise_p/100
Out[32]:
time magnitude
0 13799454.44 1.566276
1 13799454.47 14.746487
2 13799454.48 14.774490
3 13799454.49 14.748862
4 13799456.99 13.714542
As explained above, I want to create groups based on time. So expected subgroups would be rows with times 13799454and 13799456. I do this:
如上所述，我想根据时间创建组。所以预期的子组将是带有时间13799454和的行13799456。我这样做：
In [37]: ts = rise_p['time']/100
In [38]: s = rise_p/100
In [39]: new_re_df = [s.iloc[np.where(int(ts) == int(i))] for i in ts]
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-39-5ea498cf32b2> in <module>()
----> 1 new_re_df = [s.iloc[np.where(int(ts) == int(i))] for i in ts]
TypeError: only length-1 arrays can be converted to Python scalars
How do I convert tsinto an Integer Series since int() doesn't take a Series or a list as an argument? Is there any method in pandas which does this?
ts由于 int() 不将系列或列表作为参数，我如何转换为整数系列？大熊猫有什么方法可以做到这一点吗？
采纳答案by drexiya
Try converting with astype:
尝试使用 astype 进行转换：
new_re_df = [s.iloc[np.where(ts.astype(int) == int(i))] for i in ts]
Edit
编辑
On suggestion by @Rutger Kassies a nicer way would be to cast series and then groupby:
根据@Rutger Kassies 的建议，更好的方法是先投射系列，然后再分组：
rise_p['ts'] = (rise_p.time / 100).astype('int')
ts_grouped = rise_p.groupby('ts')
...
回答by Jeff
Here's a different way to solve your problem
这是解决您问题的另一种方法
In [3]: df
Out[3]:
time magnitude
0 1379945444 156.627598
1 1379945447 1474.648726
2 1379945448 1477.448999
3 1379945449 1474.886202
4 1379945699 1371.454224
In [4]: df.dtypes
Out[4]:
time int64
magnitude float64
dtype: object
Convert your epoch timestamps to seconds
将您的纪元时间戳转换为秒
In [7]: df['time'] = pd.to_datetime(df['time'],unit='s')
Set the index
设置索引
In [8]: df.set_index('time',inplace=True)
In [9]: df
Out[9]:
magnitude
time
2013-09-23 14:10:44 156.627598
2013-09-23 14:10:47 1474.648726
2013-09-23 14:10:48 1477.448999
2013-09-23 14:10:49 1474.886202
2013-09-23 14:14:59 1371.454224
Groupby 1min and mean the results (how=can be an arbitrary function as well)
Groupby 1min 并表示结果（how=也可以是任意函数）
In [10]: df.resample('1Min',how=np.mean)
Out[10]:
magnitude
time
2013-09-23 14:10:00 1145.902881
2013-09-23 14:11:00 NaN
2013-09-23 14:12:00 NaN
2013-09-23 14:13:00 NaN
2013-09-23 14:14:00 1371.454224
回答by winni2k
Here's another quite general way to convert tsto a Seriesof type int:
这是转换ts为 a Seriesof type的另一种非常通用的方法int：
rise_p['ts'] = (rise_p.time / 100).apply(lambda val: int(val))
applyallows you to apply an arbitrary function to your Seriesobject value by value. applyalso works on columns of a DataFrame object.
apply允许您Series按值将任意函数应用于对象值。apply也适用于 DataFrame 对象的列。
