package com.richardszkcs.injectjsintowebview
import android.net.Uri
import android.support.v7.app.AppCompatActivity
import android.os.Bundle
import android.webkit.JavascriptInterface
import kotlinx.android.synthetic.main.activity_main.*
import android.webkit.WebView
import android.webkit.WebViewClient
import android.widget.Toast
class MainActivity : AppCompatActivity() {
override fun onCreate(savedInstanceState: Bundle?) {
super.onCreate(savedInstanceState)
setContentView(R.layout.activity_main)
sendButton.setOnClickListener { loadWebpage() }
}
@Throws(UnsupportedOperationException::class)
fun buildUri(authority: String) : Uri {
val builder = Uri.Builder()
builder.scheme("https")
.authority(authority)
return builder.build()
}
@JavascriptInterface
fun reCaptchaCallbackInAndroid(token: String) {
val tok = token.substring(0, token.length / 2) + "..."
Toast.makeText(this.applicationContext, tok, Toast.LENGTH_LONG).show()
}
fun loadWebpage() {
webView.getSettings().setJavaScriptEnabled(true)
webView.addJavascriptInterface(this, "android")
webView.getSettings().setBuiltInZoomControls(false)
webView.loadUrl("https://richardszkcs.github.io/recaptcha-test/")
webView.webViewClient = object : WebViewClient() {
override fun onPageFinished(view: WebView, url: String) {
super.onPageFinished(view, url)
webView.loadUrl("""
javascript:(function() {
window.onCaptchaSuccess = function(token) {
android.reCaptchaCallbackInAndroid(token);
}
})()
""".trimIndent())
}
}
}
}我有上表。我想汇总每个ID的每一列。例如，我需要计算每个window.onCaptchaSuccess的有效和关闭信用的数量，然后使用计数的值为active_credits和closed_credits创建一列。与+-------+------------+---------------+-----------------+
| INDEX | SK_ID_CURR | CREDIT_ACTIVE | CREDIT_TYPE |
+-------+------------+---------------+-----------------+
| 0 | 215354 | Closed | Consumer credit |
+-------+------------+---------------+-----------------+
| 1 | 215354 | Active | Credit card |
+-------+------------+---------------+-----------------
| 2 | 215354 | Active | Consumer credit |
+-------+------------+---------------+-----------------+
| 3 | 215354 | Active | Credit card |
+-------+------------+---------------+-----------------+
| 4 | 215354 | Active | Consumer credit |
+-------+------------+---------------+-----------------+
| 5 | 215354 | Active | Credit card |
+-------+------------+---------------+-----------------+
| 6 | 215354 | Active | Consumer credit |
+-------+------------+---------------+-----------------+
| 7 | 162297 | Closed | Consumer credit |
+-------+------------+---------------+-----------------+
| 8 | 162297 | Closed | Consumer credit |
+-------+------------+---------------+-----------------+
| 9 | 162297 | Active | Credit card |
+-------+------------+---------------+-----------------+
| 10 | 162297 | Active | Credit card |
+-------+------------+---------------+-----------------+
| 11 | 162297 | Closed | Consumer credit |
+-------+------------+---------------+-----------------+
| 12 | 162297 | Active | Mortgage |
+-------+------------+---------------+-----------------+
| 13 | 402440 | Active | Consumer credit |
+-------+------------+---------------+-----------------+
| 14 | 238881 | Closed | Credit card |
+-------+------------+---------------+-----------------+
相同。
喜欢：
SK_ID_CURR答案 0 :(得分：4)
对于此数据框：
d={'SK_ID_CURR':[215354, 215354, 215354, 215354, 215354, 215354, 215354, 162297, 162297, 162297, 162297, 162297, 162297,402440 ,238881],
'CREDIT_ACTIVE':['Closed', 'Active', 'Active', 'Active', 'Active', 'Active', 'Active', 'Closed', 'Closed', 'Active', 'Active', 'Closed', 'Active', 'Active', 'Closed' ],
'CREDIT_TYPE':['Consumer credit', 'Credit card', 'Consumer credit', 'Credit card', 'Consumer credit', 'Credit card', 'Consumer credit', 'Consumer credit', 'Consumer credit', 'Credit card', 'Credit card', 'Consumer credit', 'Mortgage', 'Consumer credit', 'Credit card', ]}
df=pd.DataFrame(d)
print(df)
输出：
SK_ID_CURR CREDIT_ACTIVE CREDIT_TYPE
0 215354 Closed Consumer credit
1 215354 Active Credit card
2 215354 Active Consumer credit
3 215354 Active Credit card
4 215354 Active Consumer credit
5 215354 Active Credit card
6 215354 Active Consumer credit
7 162297 Closed Consumer credit
8 162297 Closed Consumer credit
9 162297 Active Credit card
10 162297 Active Credit card
11 162297 Closed Consumer credit
12 162297 Active Mortgage
13 402440 Active Consumer credit
14 238881 Closed Credit card
您可以尝试以下操作：
aggregations = {
'CREDIT_ACTIVE': { # work on this column,
'CREDIT_ACTIVE': lambda x: list(x).count('Active'),
'CREDIT_CLOSED': lambda x: list(x).count('Closed')
},
'CREDIT_TYPE': { # work on this column,
'CONSUMER_CREDIT ': lambda x: list(x).count('Consumer credit'),
'CREDIT_CARD': lambda x: list(x).count('Credit card')
}}
temp=df.groupby('SK_ID_CURR').agg(aggregations).reset_index()
temp.columns = [e[1] for e in temp.columns.tolist()]
print(temp)
输出：
CREDIT_ACTIVE CREDIT_CLOSED CONSUMER_CREDIT CREDIT_CARD
0 162297 3 3 3 2
1 215354 6 1 4 3
2 238881 0 1 0 1
3 402440 1 0 1 0
答案 1 :(得分：1)
另一种方式，也许有些乏味，但可能会引入一些不同的东西。
creditClosed = df[df.CREDIT_ACTIVE == 'Closed']
creditOpened = df[df.CREDIT_ACTIVE == 'Active']
creditTypeCo = df[df.CREDIT_TYPE == 'Credit card']
creditTypeCr = df[df.CREDIT_TYPE == 'Consumer credit']
a = creditClosed.groupby(['SK_ID_CURR']).agg({'CREDIT_ACTIVE':'count'}).reset_index()
b = creditOpened.groupby(['SK_ID_CURR']).agg({'CREDIT_ACTIVE':'count'}).reset_index()
c = creditTypeCo.groupby(['SK_ID_CURR']).agg({'CREDIT_TYPE':'count'}).reset_index()
d = creditTypeCr.groupby(['SK_ID_CURR']).agg({'CREDIT_TYPE':'count'}).reset_index()
ab = pd.merge(a, b, how = 'outer', on = 'SK_ID_CURR')
abc = pd.merge(ab, c, how = 'outer', on = 'SK_ID_CURR')
final = pd.merge(abc, d, how = 'outer', on = 'SK_ID_CURR')
final.rename(columns = {'CREDIT_ACTIVE_x': 'CREDIT_CLOSED', 'CREDIT_ACTIVE_y': 'CREDIT_ACTIVE', 'CREDIT_TYPE_x': 'CREDIT_CARD', 'CREDIT_TYPE_y': 'CONSUMER_CREDIT'}, inplace = True)
final.fillna(0)
输出：
CREDIT_ACTIVE CREDIT_CLOSED CONSUMER_CREDIT CREDIT_CARD
0 162297 3 3 3 2
1 215354 6 1 4 3
2 238881 0 1 0 1
3 402440 1 0 1 0
答案 2 :(得分：0)
您可以使用>$null生成伪列，例如：dummies of the dataframe
使用“ SK_ID_CURR”列将其连接起来，然后可以按“ SK_ID_CURR”进行分组。之后，使用pd.get_dummies(df.drop(columns=['SK_ID_CURR']))按总和汇总数据。
最后，这是有意义地重命名列的问题。
使用pandas在python中进行示例代码：
agg([sum])答案 3 :(得分：0)
构造一个帮助器列后，您可以加入几个pd.crosstab结果。
来自@AllaTarighati的数据。
df['TYPE'] = np.where(df['CREDIT_TYPE'].str.contains('credit', case=False, na=False),
'Credit', 'Mortgage')
cross1 = pd.crosstab(df['SK_ID_CURR'], df['TYPE'] + '_' + df['CREDIT_ACTIVE'])
cross2 = pd.crosstab(df['SK_ID_CURR'], df['CREDIT_TYPE'])
res = cross1.join(cross2)
结果
print(res)
Credit_Active Credit_Closed Mortgage_Active Consumer credit \
SK_ID_CURR
162297 2 3 1 3
215354 6 1 0 4
238881 0 1 0 0
402440 1 0 0 1
Credit card Mortgage
SK_ID_CURR
162297 2 1
215354 3 0
238881 1 0
402440 0 0