[英]Sum values from a list, selected by another list
我有球隊名稱和分數的數據:
Team A 9,
Team A 13,
Team B 24,
Team C 6,
Team A 15,
Team B 10,
Team C 19,
Team A 30,
Team B 5,
但信息存儲在 2 個列表中:
List_team = ['Team A', 'Team A', 'Team B', 'Team C',
'Team A', 'Team B', 'Team C', 'Team A', 'Team B']
List_score = [9, 13, 24, 6, 15, 10, 19, 30, 5,]
我需要 A 隊的分數和 B 隊的分數之和。
也許在壓縮 2 個列表后對它們進行排序是第一步。
最好的方法是什么?
這是使用字典的一種方法:
teams = {}
teams_and_scores = zip(List_team,List_score) #this will pair each team with their score
for team, score in teams_and_scores:
if team in teams:
teams[team] += score #add score if team is already in dictionary
else:
teams[team] = score #add team to dictionary with their score
使用您的示例:
>>> List_team = ['Team A', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B']
>>> List_score = [9, 13, 24, 6, 15, 10, 19, 30, 5,]
>>> teams = {}
>>> teams_and_scores = zip(List_team,List_score) #this will pair each team with their score
>>> for team, score in teams_and_scores:
if team in teams:
teams[team] += score #add score if team is already in dictionary
else:
teams[team] = score #add team to dictionary with their score
>>> teams
{'Team A': 67, 'Team B': 39, 'Team C': 25}
>>> teams['Team A']
67
使用defaultdict
,如果它還不存在,它將使用給定的鍵和默認值填充字典條目:
import collections
List_team = ['Team A', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B']
List_score = [9, 13, 24, 6, 15, 10, 19, 30, 5]
result = collections.defaultdict(list)
for k, s in zip(List_team, List_score):
result[k].append(s)
然后,您可以對這些數據執行任意數量的操作,例如將值傳遞給sum()
:
>>> sum(result['Team A'])
67
In [9]: result = {}
In [10]: for x in enumerate(List_team):
...: result.setdefault(x[1], 0)
...: result[x[1]] += List_score[x[0]]
...:
In [11]: result
Out[11]: {'Team A': 67, 'Team B': 39, 'Team C': 25}
或者
In [15]: result = {}
In [16]: for x, y in zip(List_team, List_score):
...: result.setdefault(x, 0)
...: result[x] += y
...:
In [17]: result
Out[17]: {'Team A': 67, 'Team B': 39, 'Team C': 25}
你可以用熊貓做到這一點:
import pandas as pd
List_team = ['Team A', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B']
List_score = [9, 13, 24, 6, 15, 10, 19, 30, 5]
df = pd.DataFrame({'Team': List_team, 'Score': List_score})
df1 = df.groupby('Team').sum()
In [47]: df1
Out[47]:
Score
Team
Team A 67
Team B 39
Team C 25
使用理解來填充設置為零的默認字典,然后按位置填充它。 其他解決方案可能更慣用,但這個解決方案似乎最人性化。
List_team = ['Team A', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B']
List_score = [9, 13, 24, 6, 15, 10, 19, 30, 5]
scores = { x : 0 for x in List_team }
for idx,team in enumerate(List_team):
scores[team] += List_score[idx]
print scores # all scores
結果:
{'Team A': 67, 'Team C': 25, 'Team B': 39}
更新:一次只獲得一支球隊的分數:
print scores['Team A'] # Just Team A
結果:
67
這是基於 TigerhawkT3 的回答,我發現最好使用 int 而不是 list 作為 defaultdict。
import collections
List_team = ['Team A', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B', 'Team C', 'Team A', 'Team B']
List_score = [9, 13, 24, 6, 15, 10, 19, 30, 5]
result = collections.defaultdict(int)
for k, s in zip(List_team, List_score):
result[k] += s
結果將返回{'Team A': 67, 'Team B': 39, 'Team C': 25}
無需在這里匯總列表。
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