出于多種原因，我想用 numpy 重寫下面的這個（作業！）代碼，但我找不到一個好的方法；#1，我以前從未使用過 numpy，而且通常是新的，#2，Python 太慢了，#3，我想使用 print(m[:,0]) 輸出一個名稱列，因為 # 4、itertools組合只輸出二維串列，不是二維numpy陣列，我不行。
def compCheck(m): # function to check how many attributes the group shares
rowsNum = len(m)
columnsNum = len(m[0])
sCount = 0 # counts the non empties in a row
matches = 0 # counts the total number of matches
for a in range(2,columnsNum):
for b in range(0,rowsNum):
if m[b][a]: # if entry isn't blank
sCount = 1
if sCount >= 3:
matches = 1
sCount = 0
print (matches)
from itertools import combinations
teamSize = 5
for i in combinations(masterList, teamSize):
compCheck(i)
為了解釋這段代碼做什么（或應該做什么），它創建了一個包含 5 行的每個唯一組合的串列，無需從 2d 串列（稱為 masterList）中替換。它查看每個組合并檢查列（偏移 2，因此不計算名稱）。如果該列中的 5 個條目中至少有 3 個被填充，那么它將將該列計為匹配項。然后它回傳匹配的總數并移動到下一個組合。
檢查示例應為：
Input: compCheck([["Alex", "Smith", "Chess", "Skiing", "", ""],
["Bob", "Dole", "Chess", "", "", ""],
["Charlie", "Chaplin", "Chess", "", "", ""],
["Daisy", "Buchanon", "", "", "", "Partying"],
["Emily", "Evans", "Chess", "Skiing", "", ""]]
Output: "1 for ['Alex' 'Bob' 'Charlie' 'Daisy' 'Emily']"
輸入示例與上面的串列大致相同（但有更多行），因此我將僅發布 6 行串列的示例：
from itertools import combinations
teamSize = 5
masterList = [["Alex", "Smith", "Chess", "Skiing", "", ""],
["Bob", "Dole", "Chess", "", "", ""],
["Charlie", "Chaplin", "Chess", "", "", ""],
["Daisy", "Buchanon", "", "", "", "Partying"],
["Emily", "Evans", "Chess", "Skiing", "", ""],
["Frank", "Ferdinand", "", "Skiing", "", ""]]
for i in combinations(masterList, teamSize):
compCheck(i)
Output: ["1 for ['Alex' 'Bob' 'Charlie' 'Daisy' 'Emily']",
"1 for ['Alex' 'Bob' 'Charlie' 'Daisy' 'Frank']",
"2 for ['Alex' 'Bob' 'Charlie' 'Emily' 'Frank']",
"2 for ['Alex' 'Bob' 'Daisy' 'Emily' 'Frank']",
"2 for ['Alex' 'Charlie' 'Daisy' 'Emily' 'Frank']",
"1 for ['Bob' 'Charlie' 'Daisy' 'Emily' 'Frank]"]
uj5u.com熱心網友回復：
我認為pandas.DataFrame在這種情況下更合適。
masterList = [
["Alex", "Smith", "Chess", "Skiing", "", ""],
["Bob", "Dole", "Chess", "", "", ""],
["Charlie", "Chaplin", "Chess", "", "", ""],
["Daisy", "Buchanon", "", "", "", "Partying"],
["Emily", "Evans", "Chess", "Skiing", "", ""],
["Frank", "Ferdinand", "", "Skiing", "", ""]
]
df = (
pd.DataFrame(
masterList,
columns = ['name','surname','chess','skiing','whatever','partying']
)
.drop(columns='surname')
.set_index('name')
.applymap(bool)
)
以下是轉換后的資料：
我們必須考慮combinations回傳一個元組序列，而不是一個二維串列。因此，我們必須在提取資料之前將每個組合轉換為串列：
NGroup = 5
minShare = 3
for combo in combinations(df.index, NGroup):
print(
'{count} for {combo}'.format(
count=(df.loc[[*combo]].sum() >= minShare).sum(),
combo=', '.join(combo)
)
)
這是輸出：
如果numpy是選擇，則此代碼可用于相同的輸出：
data = np.array(masterList)
captions = data[:, 0]
hobbies = (data[:, 2:] != '')
for combo in combinations(range(len(hobbies)), NGroup):
print(
'{count} for {combo}'.format(
count=(hobbies[[*combo]].sum(axis=0) >= minShare).sum(),
combo=', '.join(captions[[*combo]])
)
)
uj5u.com熱心網友回復：
您可以使用 numpy 輕松完成此操作
例如
def compCheck(m): # function to check how many attributes the group shares
rowsNum = len(m)
columnsNum = len(m[0])
sCount = 0 # counts the non empties in a row
matches = 0 # counts the total number of matches
for a in range(2,columnsNum):
for b in range(0,rowsNum):
if m[b][a]: # if entry isn't blank
sCount = 1
if sCount >= 3:
matches = 1
sCount = 0
return matches
from itertools import combinations
import numpy as np
teamSize = 5
masterList = [['A','B',0,1,0,1],
['A','B',0,0,0,1],
['A','B',1,1,0,0],
['A','B',0,1,1,1],
['A','B',0,1,0,0],
['A','B',1,1,0,1],
['A','B',1,1,0,0],
['A','B',0,1,1,1],
['A','B',0,0,1,1],
['A','B',1,1,0,1], ]
for i in combinations(masterList, teamSize):
Mat2D = np.array([l[2:] for l in i])
print(np.sum(np.count_nonzero(np.array(Mat2D),axis=0) >= 3))
print(compCheck(i))
如果矩陣是正確的，你說的兩個第一個值是字串。
最好直接從 masterList 中洗掉名稱
def compCheck(m): # function to check how many attributes the group shares
rowsNum = len(m)
columnsNum = len(m[0])
sCount = 0 # counts the non empties in a row
matches = 0 # counts the total number of matches
for a in range(0,columnsNum):
for b in range(0,rowsNum):
if m[b][a]: # if entry isn't blank
sCount = 1
if sCount >= 3:
matches = 1
sCount = 0
return matches
from itertools import combinations
import numpy as np
teamSize = 5
masterList = [['A','B',0,1,0,1],
['A','B',0,0,0,1],
['A','B',1,1,0,0],
['A','B',0,1,1,1],
['A','B',0,1,0,0],
['A','B',1,1,0,1],
['A','B',1,1,0,0],
['A','B',0,1,1,1],
['A','B',0,0,1,1],
['A','B',1,1,0,1], ]
masterList = np.array([l[2:] for l in masterList])
for i in combinations(masterList, teamSize):
print(np.sum(np.count_nonzero(np.array(i),axis=0) >= 3))
print(compCheck(i))
更新
這里有你的資料的代碼
from itertools import combinations
import numpy as np
teamSize = 5
masterList = [["Alex", "Smith", "Chess", "Skiing", "", ""],
["Bob", "Dole", "Chess", "", "", ""],
["Charlie", "Chaplin", "Chess", "", "", ""],
["Daisy", "Buchanon", "", "", "", "Partying"],
["Emily", "Evans", "Chess", "Skiing", "", ""],
["Frank", "Ferdinand", "", "Skiing", "", ""]]
for i in combinations(masterList, teamSize):
Mat2D = np.array([l[2:] for l in i])
check = np.sum(np.count_nonzero(np.array(Mat2D), axis=0) >= 3)
if check:
print(check, ' for ', [l[0] for l in i])
與給予
1 for ['Alex', 'Bob', 'Charlie', 'Daisy', 'Emily']
1 for ['Alex', 'Bob', 'Charlie', 'Daisy', 'Frank']
2 for ['Alex', 'Bob', 'Charlie', 'Emily', 'Frank']
2 for ['Alex', 'Bob', 'Daisy', 'Emily', 'Frank']
2 for ['Alex', 'Charlie', 'Daisy', 'Emily', 'Frank']
1 for ['Bob', 'Charlie', 'Daisy', 'Emily', 'Frank']
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