Question

我发布的内容更简单，因为我觉得它很容易理解，但是提到你的评论，我错了，所以我编辑了这个问题：

所以这是代码。我想在没有循环的情况下这样做，是否应该在熊猫中完成？

import pandas as pd

myval = [0.0,1.1, 2.2, 3.3, 4.4, 5.5,6.6, 7.7, 8.8,9.9]
s1 = [0,0,1,1,0,0,1,1,0,1]
s2 = [0,0,1,0,1,0,1,0,1,1] 

posin = [10,0,0,0,0,0,0,0,0,0] 
posout = [0,0,0,0,0,0,0,0,0,0]
sig = ['-']

d = {'myval' : myval, 's1' : s1, 's2' : s2}

d = pd.DataFrame(d)

'''
normaly the dataframe should be with the 6 col,
but I can't make the part below working in the df.(THAT is the problem !!)
The real df is 5000+ row, and this should be done for 100+ sets of values,
so this way is not eligible. Too slow.
'''

for i in xrange(1,len(myval)) :
    if (s1[i]== 1) & (s2[i] == 1) & (posin[i-1] != 0 ) :
        posin[i]= 0
        posout[i]= posin[i-1] / myval[i]
        sig.append( 'a')
    elif (s1[i] == 0) & (s2[i] == 1) & (posin[i-1] == 0) :
        posin[i]= posout[i-1] * myval[i]
        posout[i] = 0
        sig.append( 'v')
    else :
        posin[i] =  posin[i-1]
        posout[i] = posout[i-1]
        sig.append('-')

d2 = pd.DataFrame({'posin' : posin , 'posout' : posout , 'sig' : sig })

d = d.join(d2)

#the result wanted :

print d

    myval  s1  s2  posin     posout    sig
0    0.0   0   0  10.000000  0.000000   -
1    1.1   0   0  10.000000  0.000000   -
2    2.2   1   1   0.000000  4.545455   a
3    3.3   1   0   0.000000  4.545455   -
4    4.4   0   1  20.000000  0.000000   v
5    5.5   0   0  20.000000  0.000000   -
6    6.6   1   1   0.000000  3.030303   a
7    7.7   1   0   0.000000  3.030303   -
8    8.8   0   1  26.666667  0.000000   v
9    9.9   1   1   0.000000  2.693603   a

任何帮助？

谢谢!!

Answer 1

我希望以下内容可能会起作用（如评论中所述），但是（令人惊讶的是？）这种np.where的使用引发了ValueError: shape mismatch: objects cannot be broadcast to a single shape（使用1D从2D中进行选择）：

np.where(df.s1 & df.s2,
         pd.DataFrame({"bin": 0, "bout": df.bin.diff() / df.myval}),
         np.where(df.s1,
                  pd.DataFrame({"bin": df.bout.diff() * df.myval, "bout": 0}),
                  pd.DataFrame({"bin": df.bin.diff(), "bout": df.bout.diff()})))

作为使用where的替代方法，我将分阶段构建它：

res = pd.DataFrame({"bin": 0, "bout": df.bin.diff() / df.myval})
res.update(pd.DataFrame({"bin": df.bout.diff() * df.myval,
                         "bout": 0}).loc[(df.s1 == 1) & (df.s2 == 0)])
res.update(pd.DataFrame({"bin": df.bin.diff(),
                         "bout": df.bout.diff()}).loc[(df.s1 == 0) & (df.s2 == 0)])

然后你可以将它分配给df中的两列：

df[["bin", "bout"]] = res

Answer 2

代码指的是Andy Hayden的回答：

import pandas as pd

myval = [0.0,1.1, 2.2, 3.3, 4.4, 5.5,6.6, 7.7, 8.8,9.9]
s1 = [0,0,1,1,0,0,1,1,0,1]
s2 = [0,0,1,0,1,0,1,0,1,1] 

posin = [10,0,0,0,0,0,0,0,0,0] 
posout = [0,0,0,0,0,0,0,0,0,0]
sig = ['-']

d = {'myval' : myval, 's1' : s1, 's2' : s2,'posin' : posin , 'posout' : posout   }

d = pd.DataFrame(d)

res = pd.DataFrame({"posin": 10, 'sig' : '-', "posout": d.posin.diff() / d.myval})

res.update(pd.DataFrame({"posin": 0,
                         'sig' : 'a',
                         "posout":d.posin.diff() / d.myval }
                         ).loc[(d.s1 == 1) & (d.s2 == 1) & (d.posin.diff() != 0)  ])

res.update(pd.DataFrame({"posin": d.posout.diff() * d.myval,
                         'sig' : 'v',
                         "posout": 0}
                         ).loc[(d.s1 == 0) & (d.s2 == 1) & (d.posin.diff()) == 0])


d[["posin", "posout", 'sig']] = res

print d

   myval  posin  posout  s1  s2 sig
0    0.0     10       0   0   0   v
1    1.1      0       0   0   0   v
2    2.2      0       0   1   1   v
3    3.3      0       0   1   0   v
4    4.4      0       0   0   1   v
5    5.5      0       0   0   0   v
6    6.6      0       0   1   1   v
7    7.7      0       0   1   0   v
8    8.8      0       0   0   1   v
9    9.9      0       0   1   1   v

Python Pandas：如何同时设置2列？

2 个答案: