Python版本:2.7.6 Numpy版本:1.10.2 熊猫:0.17.1
我知道.ix现已弃用,但我正在使用遗留系统并看到这种行为与.ix和我预先准备好了
# Native Python List Indexing is exclusive on the end index
[0, 1, 2, 3][0:1] # returns [0] indexes with [0, 1)
# Native Numpy
import numpy as np
numpyArray = np.reshape(np.arange(4), (2, 2))
numpyArray[0:1, 0:1] # returns array([[0]]), indexes with [0, 1) in rows and [0, 1) in columns
####### Pandas #######
import pandas as pd
dataFrame = pd.DataFrame(numpyArray)
# Pandas with iloc #
dataFrame.iloc[0:1, 0:1] # returns 0, indexes with [0, 1) in rows and [0, 1) in columns
# Pandas with ix #
dataFrame.ix[0:1, 0:1] # returns [[0, 1], [2, 3] indexes with [0, 1] in rows and [0, 1] in columns
答案 0 :(得分:4)
.ix
是基于标签的索引(与.loc
相同),文档状态包括与iloc
不同的停止范围值,这是开放闭合范围所以不是包括停止范围值,这是设计
这样做的原因是因为如果您的索引是例如字符串,那么选择一个您不知道结束范围值应该是什么的范围会有问题:
In[274]:
df = pd.DataFrame(np.random.randn(5,3), columns=list('abc'), index=list('vwxyz'))
df
Out[274]:
a b c
v -0.488627 0.213183 0.224104
w -0.200328 -1.138937 0.815568
x -1.131868 -0.562758 0.088719
y 0.120701 -0.863737 0.246295
z -0.808140 0.253376 0.645974
In[275]:
df.ix['w':'y']
Out[275]:
a b c
w -0.200328 -1.138937 0.815568
x -1.131868 -0.562758 0.088719
y 0.120701 -0.863737 0.246295
如果它没有包含最后一行的结束值,您需要知道必须传递'z'
才能在'z'
之前返回标签以获得上述结果
<强>更新强>
请注意0.20.1
已弃用loc
,您应使用var user = new Parse.User();
var base64 = "V29ya2luZyBhdCBQYXJzZSBpcyBncmVhdCE=";
var imageFile = new Parse.File("image.png", { base64: base64 });
user.set("username", $scope.user.username);
user.set("password", $scope.user.password);
user.set("email", $scope.user.email);
user.set("picture", imageFile);//getting file need url
user.signUp(null, {
success: function (user) {
$state.go('app.home');
},
error: function (user, error) {
alert("Error: " + error.code + " " + error.message);
}
});