我有一个.csv文件：
20376.65,22398.29,4.8,0.0,1.0,2394.0,6.1,89.1,0.0,4.027,9.377,0.33,0.28,0.36,51364.0,426372.0,888388.0,0.0,2040696.0,57.1,21.75,25.27,0.0,452.0,1046524.0,1046524.0,1
7048.842,8421.754,1.44,0.0,1.0,2394.0,29.14,69.5,0.0,4.027,9.377,0.33,0.28,0.36,51437.6,426964.0,684084.0,0.0,2040696.0,57.1,12.15,14.254,3.2,568.8,1046524.0,1046524.0,1
3716.89,4927.62,0.12,0.0,1.0,2394.0,26.58,73.32,0.0,4.027,9.377,0.586,1.056,3.544,51456.0,427112.0,633008.0,0.0,2040696.0,57.1,9.75,11.5,4.0,598.0,1046524.0,1046524.0,1
3716.89,4927.62,0.0,0.0,1.0,2394.0,17.653333333,82.346666667,0.0,4.027,9.377,0.84066666667,1.796,5.9346666667,51487.2,427268.0,481781.6,0.0,2040696.0,57.1,9.75,11.5,4.0,598.0,1046524.0,1046524.0,1
3716.89,4927.62,0.0,0.0,1.0,2394.0,16.6,83.4,0.0,4.027,9.377,0.87,1.88,6.18,51492.0,427292.0,458516.0,0.0,2040696.0,57.1,9.75,11.5,4.0,598.0,1046524.0,1046524.0,1
我正在使用pandas数据帧对其进行规范化，但我在.csv文件中缺少值：
.703280701968,0.867283950617,,,,0.0971635485818,-0.132770066385,,0.318518516666,-inf,-0.742913580247,-0.74703196347,-0.779350940252,-0.659592176966,-0.483438485804,0.565758716954,,,-inf,-0.274046377081,0.705774765311,-0.281481481478,-0.596841230258,,,1
0.104027493068,-0.0493827160494,,,,0.0199155099578,-0.0175015087508,,0.318518516666,-inf,-0.401580246914,-0.392694063927,-0.331530968381,-0.401165210674,-0.337539432177,0.426956186355,,,-inf,-0.373755558635,-0.294225234689,0.518518518522,-0.232751454697,,,1
0.104027493068,-0.132716049383,,,,-0.2494467914,0.254878294116,,0.318518516666,-inf,-0.0620246913541,-0.0547945205479,0.00470906912955,0.0370370365169,-0.183753943218,0.0159880797389,,,-inf,-0.373755558635,-0.294225234689,0.518518518522,-0.232751454697,,,1
0.104027493068,-0.132716049383,,,,-0.281231140616,0.286662643331,,0.318518516666,-inf,-0.0229135802474,-0.0164383561644,0.0392144605923,0.104452766854,-0.160094637224,-0.0472377828174,,,-inf,-0.373755558635,-0.294225234689,0.518518518522,-0.232751454697,,,1
0.104027493068,-0.132716049383,,,,-0.566083283042,0.571514785757,,0.318518516666,-inf,0.201086419753,0.199086757991,0.184362139917,0.104452766854,-0.160094637224,-0.0472377828174,,,-inf,-0.373755558635,-0.294225234689,0.518518518522,-0.232751454697,,,1
我的代码：
import pandas as pd
df = pd.read_csv('pooja.csv',index_col=False)
df_norm = (df.ix[:, 1:-1] - df.ix[:, 1:-1].mean()) / (df.ix[:, 1:-1].max() - df.ix[:, 1:-1].min())
rslt = pd.concat([df_norm, df.ix[:,-1]], axis=1)
rslt.to_csv('example.csv',index=False,header=False)
代码有什么问题？为什么.csv文件中缺少值？
答案 0 :(得分：1)
您获得了很多NaN，因为将0除以0。见broadcasting behaviour。更好的解释是here。
我使用之前question中的代码，因为我认为没有必要使用df.ix[:, 1:-1]进行切片。在使用切片进行标准化后，我得到空DataFrame。
import pandas as pd
import numpy as np
import io
temp=u"""20376.65,22398.29,4.8,0.0,1.0,2394.0,6.1,89.1,0.0,4.027,9.377,0.33,0.28,0.36,51364.0,426372.0,888388.0,0.0,2040696.0,57.1,21.75,25.27,0.0,452.0,1046524.0,1046524.0,1
7048.842,8421.754,1.44,0.0,1.0,2394.0,29.14,69.5,0.0,4.027,9.377,0.33,0.28,0.36,51437.6,426964.0,684084.0,0.0,2040696.0,57.1,12.15,14.254,3.2,568.8,1046524.0,1046524.0,1
3716.89,4927.62,0.12,0.0,1.0,2394.0,26.58,73.32,0.0,4.027,9.377,0.586,1.056,3.544,51456.0,427112.0,633008.0,0.0,2040696.0,57.1,9.75,11.5,4.0,598.0,1046524.0,1046524.0,1
3716.89,4927.62,0.0,0.0,1.0,2394.0,17.653333333,82.346666667,0.0,4.027,9.377,0.84066666667,1.796,5.9346666667,51487.2,427268.0,481781.6,0.0,2040696.0,57.1,9.75,11.5,4.0,598.0,1046524.0,1046524.0,1
3716.89,4927.62,0.0,0.0,1.0,2394.0,16.6,83.4,0.0,4.027,9.377,0.87,1.88,6.18,51492.0,427292.0,458516.0,0.0,2040696.0,57.1,9.75,11.5,4.0,598.0,1046524.0,1046524.0,1"""
#after testing replace io.StringIO(temp) to filename
df = pd.read_csv(io.StringIO(temp),index_col=None, header=None)
#print df
#filter only first 5 columns for testing
df = df.iloc[:, :5]
print df
0 1 2 3 4
0 20376.650 22398.290 4.80 0 1
1 7048.842 8421.754 1.44 0 1
2 3716.890 4927.620 0.12 0 1
3 3716.890 4927.620 0.00 0 1
4 3716.890 4927.620 0.00 0 1
#get max values by columns
print df.max()
0 20376.65
1 22398.29
2 4.80
3 0.00
4 1.00
dtype: float64
#get min values by columns
print df.min()
0 3716.89
1 4927.62
2 0.00
3 0.00
4 1.00
dtype: float64
#difference, you get 0
print (df.max() - df.min())
0 16659.76
1 17470.67
2 4.80
3 0.00
4 0.00
dtype: float64
print df - df.mean()
0 1 2 3 4
0 12661.4176 13277.7092 3.528 0 0
1 -666.3904 -698.8268 0.168 0 0
2 -3998.3424 -4192.9608 -1.152 0 0
3 -3998.3424 -4192.9608 -1.272 0 0
4 -3998.3424 -4192.9608 -1.272 0 0
#you get NaN, because divide columns 3 and 4 filled 0 to difference with index 3,4 filled 0
df_norm = (df - df.mean()) / (df.max() - df.min())
print df_norm
0 1 2 3 4
0 0.76 0.76 0.735 NaN NaN
1 -0.04 -0.04 0.035 NaN NaN
2 -0.24 -0.24 -0.240 NaN NaN
3 -0.24 -0.24 -0.265 NaN NaN
4 -0.24 -0.24 -0.265 NaN NaN
上次生成to_csv时，请从NaN ""获取，因为参数na_rep的默认值为""：
print df_norm.to_csv(index=False, header=False, na_rep="")
0.76,0.76,0.735,,
-0.04,-0.04,0.035,,
-0.24,-0.24,-0.24,,
-0.24,-0.24,-0.265,,
-0.24,-0.24,-0.265,,
如果您更改na_rep的值：
#change na_rep to * for testing
print df_norm.to_csv(index=False, header=False, na_rep="*")
0.76,0.76,0.735,*,*
-0.04,-0.04,0.035,*,*
-0.24,-0.24,-0.24,*,*
-0.24,-0.24,-0.265,*,*
-0.24,-0.24,-0.265,*,*