Question

有没有办法创建新列，表示包含两个日期时间之间的增量的各个月？输出可能是每个新月度列的二进制值。我在想这样的事情（它不起作用）：

for i in [1, 2, 3, 4, 5]:
    i_name = str(i)
    values = example['end'] - example['start'] #Example line - need to expose values here) 
    example[i_name] = values

离开这个：

    end         name        start
0   28/02/2012  joe bloggs  01/01/2012
1   15/03/2012  jane bloggs 01/02/2012
2   17/05/2012  jim bloggs  01/04/2012
3   18/04/2012  john bloggs 01/02/2012

对此：

    end         1   2   3   4   5   name        start
0   28/02/2012  1   1   0   0   0   joe bloggs  01/01/2012
1   15/03/2012  0   1   1   0   0   jane bloggs 01/02/2012
2   17/05/2012  0   0   0   1   1   jim bloggs  01/04/2012
3   18/04/2012  0   1   1   1   0   john bloggs 01/02/2012

Answer 1

我认为你可以主要使用get_dummies与stack：

#convert columns to datetime
df['end'] = pd.to_datetime(df.end, dayfirst=True)
df['start'] = pd.to_datetime(df.start, dayfirst=True)
#print df

#get months to Series
end = df['end'].dt.month
start = df['start'].dt.month

#create difference DataFrame
df1 = pd.DataFrame({'end':end, 'start':start})
        .apply(lambda x: pd.Series(range(x.start, x.end + 1)), axis=1)
print df1
     0    1    2
0  1.0  2.0  NaN
1  2.0  3.0  NaN
2  4.0  5.0  NaN
3  2.0  3.0  4.0

#create indicator variables, sum values by index
df1 = pd.get_dummies(df1.stack().reset_index(level=1, drop=True)) 
        .groupby(level=0).sum().astype(int)

#convert float columns names to int
df1.columns = df1.columns.to_series().astype(int)
print df1
   1  2  3  4  5
0  1  1  0  0  0
1  0  1  1  0  0
2  0  0  0  1  1
3  0  1  1  1  0

#append to original DataFrame
print pd.concat([df, df1], axis=1)
         end         name      start  1  2  3  4  5
0 2012-02-28   joe bloggs 2012-01-01  1  1  0  0  0
1 2012-03-15  jane bloggs 2012-02-01  0  1  1  0  0
2 2012-05-17   jim bloggs 2012-04-01  0  0  0  1  1
3 2012-04-18  john bloggs 2012-02-01  0  1  1  1  0

Answer 2

这样可行：

example = pd.read_csv(FILE, parse_dates=[0, 2], dayfirst=True)
for i in [1, 2, 3, 4, 5]:
    i_name = str(i)
    example[i_name] = example.apply(lambda example: example["start"] <= pd.datetime(2012, i, 1) <= example["end"], axis=1).astype(int)

Answer 3

首先，您必须使用pd.to_datetime将日期列转换为日期时间：

import pandas as pd
example['end'] = pd.to_datetime(example['end'], dayfirst=True) #default is ydm...
example['start'] = pd.to_datetime(example['start'], dayfirst=True)

然后在你的for循环中你可以设置适当的值：

example[str(i)] = 0
example[str(i)][( i >= example['start'].dt.month) & (example['end'].dt.month >= i)] = 1

（从jezrael的回答中窃取dt.month）导致：

import pandas as pd
example['end'] = pd.to_datetime(example['end'], dayfirst=True) #default is ydm...
example['start'] = pd.to_datetime(example['start'], dayfirst=True)

for i in range(1,13):
  example[str(i)] = 0
  example[str(i)][( i >= example['start'].dt.month) & (example['end'].dt.month >= i)] = 1

然后导致：

In[101]: example
Out[101]: 
         end         name      start  1  2  3  4  5  6  7  8  9  10  11  12
0 2012-02-28   joe bloggs 2012-01-01  1  1  0  0  0  0  0  0  0   0   0   0
1 2012-03-15  jane bloggs 2012-02-01  0  1  1  0  0  0  0  0  0   0   0   0
2 2012-05-17   jim bloggs 2012-04-01  0  0  0  1  1  0  0  0  0   0   0   0
3 2012-04-18  john bloggs 2012-02-01  0  1  1  1  0  0  0  0  0   0   0   0

你能在熊猫中分出日期时间吗？

3 个答案: