目前，我做
df.groupby('A')['B'].agg({'total' : np.sum})
以便聚合列的列名为total。但最近我收到了警告
FutureWarning: using a dict on a Series for aggregation is deprecated and will be removed in a future version
那么现在首选的方式是什么？
答案 0 :(得分：2)
我认为您需要 try {
RequestQueue requestQueue = Volley.newRequestQueue(this);
String URL = "url";
final String requestBody = "mutation M {" + "logIn" + "(" +
"countryCode:" + "\"" + "1" + "\"" +
"deviceType:" + "\"" + "Android" + "\"" +
"osVerison:" + "\"" + "5.1.1" + "\"" +
"appVerison:" + "\"" + "2.9" + "\"" +
"currentCountry:" + "\"" + "IND" + "\"" +
"language:" + "\"" + "ENG" + "\"" +
"deviceModal:" + "\"" + "ASUS" + "" +
"\")" + "{_id,countryCode,phoneNumber,emailAddress,profilePic}";
StringRequest stringRequest = new StringRequest(Request.Method.POST, URL, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Log.e("VOLLEY", response);
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e("VOLLEY", error.toString());
}
}) {
@Override
public Map<String, String> getHeaders() throws AuthFailureError {
Map<String, String> params = new HashMap<String, String>();
params.put("Cache-Control","no-cache");
params.put("Content-type", "application/graphql");
return params;
}
@Override
public byte[] getBody() throws AuthFailureError {
try {
return requestBody == null ? null : requestBody.getBytes("utf-8");
} catch (UnsupportedEncodingException uee) {
VolleyLog.wtf("Unsupported Encoding while trying to get the bytes of %s using %s", requestBody, "utf-8");
return null;
}
}
@Override
protected Response<String> parseNetworkResponse(NetworkResponse response) {
String responseString = "";
if (response != null) {
responseString = response.toString();
// can get more details such as response.headers
}
return Response.success(responseString, HttpHeaderParser.parseCacheHeaders(response));
}
};
requestQueue.add(stringRequest);
} catch (Exception e) {
e.printStackTrace();
}
或reset_index中定义rename参数：
同时检查deprecate groupby agg with a dictionary when renaming。
<强> 1
name<强> 2
df = df.groupby('A')['B'].sum().reset_index(name='total')
第3
df = df.groupby('A', as_index=False)['B'].sum().rename(columns={'B':'total'})
答案 1 :(得分：2)
以下是一些方法
In [57]: df.groupby('A')['B'].sum().to_frame('total')
Out[57]:
total
A
a 2
b 1
In [58]: df.groupby('A')['B'].agg(np.sum).to_frame('total') # or agg('sum')
Out[58]:
total
A
a 2
b 1
In [59]: df.groupby('A').agg({'B': np.sum}).rename(columns={'B': 'total'})
Out[59]:
total
A
a 2
b 1