我正在尝试将数据附加到日志中,其中列的顺序不是按字母顺序排列,但具有逻辑意义,例如
Org_Goals_1 Calc_Goals_1 Diff_Goals_1 Org_Goals_2 Calc_Goals_2 Diff_Goals_2
我正在基于不同变量进行多次计算,并通过在每次运行后附加值的字典来记录结果。有没有办法阻止df.append()函数按字母顺序排列?
答案 0 :(得分:4)
实际上,我发现了#34;高级索引"工作得很好
df2=df.ix[:,'order of columns']
答案 1 :(得分:3)
似乎你必须在追加操作后重新排序列:
In [25]:
# assign the appended dfs to merged
merged = df1.append(df2)
# create a list of the columns in the order you desire
cols = list(df1) + list(df2)
# assign directly
merged.columns = cols
# column order is now as desired
merged.columns
Out[25]:
Index(['Org_Goals_1', 'Calc_Goals_1', 'Diff_Goals_1', 'Org_Goals_2', 'Calc_Goals_2', 'Diff_Goals_2'], dtype='object')
示例:
In [26]:
df1 = pd.DataFrame(columns=['Org_Goals_1','Calc_Goals_1','Diff_Goals_1'], data = randn(5,3))
df2 = pd.DataFrame(columns=['Org_Goals_2','Calc_Goals_2','Diff_Goals_2'], data=randn(5,3))
merged = df1.append(df2)
cols = list(df1) + list(df2)
merged.columns = cols
merged
Out[26]:
Org_Goals_1 Calc_Goals_1 Diff_Goals_1 Org_Goals_2 Calc_Goals_2 \
0 0.028935 NaN -0.687143 NaN 1.528579
1 0.943432 NaN -2.055357 NaN -0.720132
2 0.035234 NaN 0.020756 NaN 1.556319
3 1.447863 NaN 0.847496 NaN -1.458852
4 0.132337 NaN -0.255578 NaN -0.222660
0 NaN 0.131085 NaN 0.850022 NaN
1 NaN -1.942110 NaN 0.672965 NaN
2 NaN 0.944052 NaN 1.274509 NaN
3 NaN -1.796448 NaN 0.130338 NaN
4 NaN 0.961545 NaN -0.741825 NaN
Diff_Goals_2
0 NaN
1 NaN
2 NaN
3 NaN
4 NaN
0 0.727619
1 0.022209
2 -0.350757
3 1.116637
4 1.947526
列的相同alpha排序也会发生在concat中,因此看起来你必须在追加后重新排序。
修改强>
另一种方法是使用join
:
In [32]:
df1.join(df2)
Out[32]:
Org_Goals_1 Calc_Goals_1 Diff_Goals_1 Org_Goals_2 Calc_Goals_2 \
0 0.163745 1.608398 0.876040 0.651063 0.371263
1 -1.762973 -0.471050 -0.206376 1.323191 0.623045
2 0.166269 1.021835 -0.119982 1.005159 -0.831738
3 -0.400197 0.567782 -1.581803 0.417112 0.188023
4 -1.443269 -0.001080 0.804195 0.480510 -0.660761
Diff_Goals_2
0 -2.723280
1 2.463258
2 0.147251
3 2.328377
4 -0.248114
答案 2 :(得分:0)
正如我所看到的,订单已丢失,但在追加时,原始数据应该具有正确的顺序。为了保持这一点,假设Dataframe' alldata'和数据框作为附加数据' newdata',附加并保持列顺序,如< alldata'将是:
alldata.append(newdata)[list(alldata)]
(我在命名的日期字段中遇到了这个问题,其中'月'将在' Minute'' Second')之间进行排序。