我正在使用商业机构数据集。这是一个宽格式的小组,每年的就业人数,比如2005年,2006年,2007年等。有一个变量,一年的业务搬到了一个新的位置,比如2006年。我想创建一个变量for移动年度的具体就业 - 也就是说,如果移动年份为x,则查找第x年的就业价值。
理想情况下,我会对此进行矢量化。这就是我现在所拥有的,但我担心索引不够通用/可能是危险的,我可能会在实际数据中得到意想不到的结果。
import pandas as pd
import numpy as np
np.random.seed(43)
## prep mock data
N = 100
industry = ['utilities','sales','real estate','finance']
city = ['sf','san mateo','oakland']
move = np.arange(2006,2010)
ind = np.random.choice(industry, N)
cty = np.random.choice(city, N)
moveyr = np.random.choice(move, N)
## place it in dataframe
jobs06 = np.random.randint(low=1,high=250,size=N)
jobs06 = np.random.randint(low=1,high=250,size=N)
jobs07 = np.random.randint(low=1,high=250,size=N)
jobs08 = np.random.randint(low=1,high=250,size=N)
jobs09 = np.random.randint(low=1,high=250,size=N)
df_city =pd.DataFrame({'industry':ind,'city':cty,'moveyear':moveyr,'jobs06':jobs06,'jobs07':jobs07,'jobs08':jobs08,'jobs09':jobs09})
df_city.head()
提供此数据:
+---+------------+------------+--------+--------+--------+--------+----------+
| | city | industry | jobs06 | jobs07 | jobs08 | jobs09 | moveyear |
+---+------------+------------+--------+--------+--------+--------+----------+
| 0 | sf | utilities | 206 | 82 | 192 | 236 | 2009 |
| 1 | oakland | utilities | 10 | 244 | 2 | 7 | 2007 |
| 2 | san mateo | finance | 182 | 164 | 49 | 66 | 2006 |
| 3 | oakland | sales | 27 | 228 | 33 | 169 | 2007 |
| 4 | san mateo | sales | 24 | 24 | 127 | 165 | 2007 |
+---+------------+------------+--------+--------+--------+--------+----------+
如果我做这样的事情,我会得到一些似乎是正确的东西,至少在这个玩具示例中,但我不肯定这是a)安全,索引,b)'正确'的pythonic方式(以及无论如何大熊猫相当于那个词。)
df_city['moveyearemp']=0 ## seemingly must declare first
for count, row in df_city.head(5).iterrows():
get_moveyear_emp = 'jobs' + str(row['moveyear'])[2:]
## is this 'proper' indexing?
df_city.ix[count,'moveyearemp'] = df_city.ix[count,get_moveyear_emp]
print df_city['moveyearemp'].head()
这确实给出了预期的结果 - 例如,236确实是2009年第一排/业务的就业;第二排等于2007年的244同上。
0 236
1 244
2 182
3 228
4 24
Name: moveyearemp, dtype: int64
答案 0 :(得分:2)
我可能会迭代这些年(因为年数少于行数):
In [11]: df_city.moveyear.unique()
Out[11]: array([2009, 2007, 2006, 2008])
这是一种方法,但我不认为我会称之为 pandastic ......
g = df_city.groupby('moveyear')
df_city['moveyearemp'] = 0
for year, ind in g.indices.iteritems():
year_abbr = str(year)[2:]
df_city.loc[ind, 'moveyearemp'] = df_city.loc[ind, 'jobs%s' % year_abbr]
你得到:
In [21]: df_city.head()
Out[21]:
city industry jobs06 jobs07 jobs08 jobs09 moveyear moveyearemp
0 sf utilities 206 82 192 236 2009 236
1 oakland utilities 10 244 2 7 2007 244
2 san mateo finance 182 164 49 66 2006 182
3 oakland sales 27 228 33 169 2007 228
4 san mateo sales 24 24 127 165 2007 24
答案 1 :(得分:0)
如果您预先计算moveyearemp
数据框(按年份编制索引的数据集),您将可以执行df_city.join(moveyearemp, on='year')