Python 按列减去元素(以熊猫为单位)
我有以下数据帧:Python 按列减去元素(以熊猫为单位),python,pandas,Python,Pandas,我有以下数据帧: frame=pd.DataFrame({"col1":[1,5,9,4,7,3],"col2":[5,8,7,9,3,4],"col3":[3,4,2,7,9,1], "col4":[2,4,7,4,9,0],"col5":[3,4,5,2,1,1],"col6":[8,7,5,4,1,2]}) 它会产生以下输出: col1 col2 col3 col4 col5 col6 0 1 5 3
frame=pd.DataFrame({"col1":[1,5,9,4,7,3],"col2":[5,8,7,9,3,4],"col3":[3,4,2,7,9,1],
"col4":[2,4,7,4,9,0],"col5":[3,4,5,2,1,1],"col6":[8,7,5,4,1,2]})
它会产生以下输出:
col1 col2 col3 col4 col5 col6
0 1 5 3 2 3 8
1 5 8 4 4 4 7
2 9 7 2 7 5 5
3 4 9 7 4 2 4
4 7 3 9 9 1 1
5 3 4 1 0 1 2
我想创建一个新的数据框架,将col1和col2、col3和col4以及col5和col6区别开来
预期产出如下:
col1-col2 col3-col4 col5-col6
0 -4 1 -5
1 -3 0 -3
2 2 -5 0
3 -5 3 -2
4 4 0 0
5 -1 1 -1
提前谢谢
df = pd.DataFrame(frame.apply(lambda x: [x['col1']-x['col2'],x['col3']-x['col4'],x['col5']-x['col6']],axis=1).tolist())
df.rename({0:'col1-col2',1:'col3-col4',2:'col4-col5'},axis=1)
如果许多列使用通用解决方案-选择“对列”和“取消对列”,请转换为numpy数组,并由contructor创建新的
DataFrame
:
#pandas 0.24+
arr = frame.iloc[:, ::2].to_numpy() - frame.iloc[:, 1::2].to_numpy()
#pandas below
#arr = frame.iloc[:, ::2].values - frame.iloc[:, 1::2].values
c = [f'{a}-{b}' for a, b in zip(frame.columns[::2], frame.columns[1::2])]
df = pd.DataFrame(arr, columns=c)
print (df)
col1-col2 col3-col4 col5-col6
0 -4 1 -5
1 -3 0 -3
2 2 -5 0
3 -5 3 -2
4 4 0 0
5 -1 1 -1
如果性能很重要,请首先转换为numpy数组,存储为变量,然后索引:
#pandas 0.24+
arr = frame.to_numpy()
#pandas below
#arr = frame.values
c = [f'{a}-{b}' for a, b in zip(frame.columns[::2], frame.columns[1::2])]
df = pd.DataFrame(arr[:, ::2] - arr[:, 1::2], columns=c)
@yatu-ya,double.values vs only only only,应该很有趣,看看性能上的差异……是的,很有趣
。values
非常昂贵。有一个很好的改善(200us)。尼斯+1
#pandas 0.24+
arr = frame.to_numpy()
#pandas below
#arr = frame.values
c = [f'{a}-{b}' for a, b in zip(frame.columns[::2], frame.columns[1::2])]
df = pd.DataFrame(arr[:, ::2] - arr[:, 1::2], columns=c)
dfr = pd.DataFrame({'col1-col2': frame.col1 - frame.col2,
'col3-col4': frame.col3 - frame.col4,
'col5-col6': frame.col5 - frame.col6})