Question

我必须将数字ID转换为我知道维度的矩阵i的索引j和M。表达式如下：

s = shape(M)
j = (ID - 1) % s[0]
i = np.int((ID - 1 - j) / s[0])

我必须保存来自数据框df的值，每种类型我都会重复以下内容：

M = np.zeros((m, n))
s = shape(M)
for k in df.index:
    ID = df['Id'][k]
    y = (ID - 1) % s[0]
    x = np.int((ID - 1 - y) / s[0])
    M[x,y] = M[x,y] + df['R'][k]

我想知道有一种方法可以避免每次循环并节省计算时间。

Answer 1

基本上，您可以M Series R MultiIndex i使用j，.unstack()使用import numpy as np import pandas as pd M = pd.DataFrame(index=range(4), columns=range(4)) s = M.shape df = pd.DataFrame({ 'ID': np.arange(4*4), 'R': np.random.rand(4*4), }) df['j'] = df.ID % s[0] df['i'] = np.floor(df.ID / s[0]).astype(np.int) #In [26]: q.df Out[26]: # ID R j i # 0 0 0.847475 0 0 # 1 1 0.343867 1 0 # 2 2 0.806913 2 0 # 3 3 0.105036 3 0 # 4 4 0.170887 0 1 # 5 5 0.782607 1 1 # 6 6 0.377907 2 1 # 7 7 0.586738 3 1 # 8 8 0.895262 0 2 # 9 9 0.290358 1 2 # 10 10 0.109228 2 2 # 11 11 0.423267 3 2 # 12 12 0.454464 0 3 # 13 13 0.604296 1 3 # 14 14 0.603850 2 3 # 15 15 0.974695 3 3 df.set_index(['i', 'j'], inplace=True) M = df['R'].unstack() # In [30]: q.M # Out[30]: # j 0 1 2 3 # i # 0 0.975895 0.506286 0.055459 0.551988 # 1 0.846078 0.114153 0.981231 0.681130 # 2 0.142864 0.050597 0.323655 0.918745 # 3 0.734922 0.951795 0.209542 0.547859直接获取矩阵jQuery(function ($) { loading = false; $('#menu_left').on('click', 'a', function (e) { e.preventDefault(); if (!loading) { loading = true; var data = { action: 'glevouir_ajax_load', nonce: glevouirajaxload.nonce, pageid: $(this).parent().attr("id") }; $.post(glevouirajaxload.url, data, function (res) { if (res.success) { $("#content").slideUp("fast", function () { $("#content").html(res.data); $(this).slideDown("fast", function () { // Animation complete. }); }); } }); loading = false; } }); }); }}：

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Answer 2

在我按照您的程序后，我的数据框df。

df.head(10):
        Old1   ID   R
i   j           
11  36  1169    0   0.0
12  32  1268    1   0.0
    34  1270    2   0.0
    35  1271    3   2.0
    36  1272    4   0.0
    37  1273    5   0.0
    40  1276    6   0.0
    41  1277    7   0.0
13  32  1371    8   7.0
    33  1372    9   0.0

是否可以将这些值与矩阵相关联，即

M[12][35]=2.0  
M[13][32]=7.0

等等。

这就是我所做的：

tmp = list(dftmp.index)
tmp = pd.DataFrame(tmp)
tmp.columns = ('a','b')
M[tmp.a, tmp.b] = dftmp.I

Python：如何将ID映射到矩阵索引？

2 个答案: