Question

如何编写嵌套的db.collection.aggregate([ { $project: { id: 1, types: { $map: { input: "$types", as: "type", in: { type: "$$type", applications: { $filter: { input: "$applications", as: "application", cond: { $allElementsTrue: { $map: { input: "$$application", in: { $eq: [ "$$this", "$$type" ] } } } } } } } } } } }, { $addFields: { types: { $map: { input: "$types", in: { $mergeObjects: [ "$$this", { count: { $reduce: { input: "$$this.applications", initialValue: 0, in: { $add: [ "$$value", { $size: "$$this" } ] } } } } ] } } } } } ])循环以访问for中一行的其他每一行？

我正在尝试在pandas.dataframe中的行之间执行一些操作我的示例代码的操作是计算每一行与另一行之间的欧几里得距离。然后将结果保存到表单中的某个列表中 pandas.dataframe。

我了解如何使用[(row_reference, name, dist)]访问pandas.dataframe中的每一行，但是我不确定如何相对于当前行访问其他每一行以执行行间操作。 / p>

df.itterrows()

我希望对当前行/索引的所有行执行一些操作import pandas as pd import numpy import math df = pd.DataFrame([{'name': "Bill", 'c1': 3, 'c2': 8}, {'name': "James", 'c1': 4, 'c2': 12}, {'name': "John", 'c1': 12, 'c2': 26}]) #Euclidean distance function where x1=c1_row1 ,x2=c1_row2, y1=c2_row1, #y2=c2_row2 def edist(x1, x2, y1, y2): dist = math.sqrt(math.pow((x1 - x2),2) + math.pow((y1 - y2),2)) return dist # Calculate Euclidean distance for one row (e.g. Bill) against each other row # (e.g. "James" and "John"). Save results to a list (N_name, dist). all_results = [] for index, row in df.iterrows(): results = [] # secondary loop to look for OTHER rows with respect to the current row # results.append(row2['name'],edist()) all_results.append(row,results)。

我希望循环执行以下操作：

edist()

具有以下预期结果输出：

In[1]:
result = []
result.append(['James',edist(3,4,8,12)])
result.append(['John',edist(3,12,8,26)])
results_all=[]
results_all.append([0,result])
result2 = []
result2.append(['John',edist(4,12,12,26)])
result2.append(['Bill',edist(4,3,12,8)])
results_all.append([1,result2])
result3 = []
result3.append(['Bill',edist(12,3,26,8)])
result3.append(['James', edist(12,4,26,12)])
results_all.append([2,result3])
results_all

Answer 1

如果数据不太长，可以检出scipy的distance_matrix：

all_results = pd.DataFrame(distance_matrix(df[['c1','c2']],df[['c1','c2']]),
                           index=df['name'],
                           columns=df['name'])

输出：

name        Bill      James       John
name                                  
Bill    0.000000   4.123106  20.124612
James   4.123106   0.000000  16.124515
John   20.124612  16.124515   0.000000

Answer 2

考虑shift并避免任何行循环。而且，因为您运行的是简单的算术运算，所以可以借助numpy进行矢量化计算，直接在列上运行表达式。

import numpy as np

df = (df.assign(c1_shift = lambda x: x['c1'].shift(1),
                c2_shift = lambda x: x['c2'].shift(1))
     )

df['dist'] = np.sqrt(np.power(df['c1'] - df['c1_shift'], 2) + 
                     np.power(df['c2'] - df['c2_shift'], 2))

print(df)
#     name  c1  c2  c1_shift  c2_shift       dist
# 0   Bill   3   8       NaN       NaN        NaN
# 1  James   4  12       3.0       8.0   4.123106
# 2   John  12  26       4.0      12.0  16.124515

是否希望每个行彼此组合，考虑自身的交叉连接并查询反向重复项：

df = (pd.merge(df.assign(key=1), df.assign(key=1), on="key")
        .query("name_x < name_y")
        .drop(columns=['key'])
     )

df['dist'] = np.sqrt(np.power(df['c1_x'] - df['c1_y'], 2) +
                     np.power(df['c2_x'] - df['c2_y'], 2))

print(df)    
#   name_x  c1_x  c2_x name_y  c1_y  c2_y       dist
# 1   Bill     3     8  James     4    12   4.123106
# 2   Bill     3     8   John    12    26  20.124612
# 5  James     4    12   John    12    26  16.124515

如何在pandas.dataframe中执行行间操作

2 个答案: