我在熊貓數據框中有一些布爾變量，我需要獲取所有唯一的元組。所以我的想法是創建一個新的變量連接值列，然后使用 pandas.DataFrame.unique() 來獲取所有唯一的元組。所以我的想法是使用二進制開發進行連接。例如，對于數據框：import pandas as pddf = pd.DataFrame({'v1':[0,1,0,0,1],'v2':[0,0,0,1,1], 'v3':[0,1,1,0,1], 'v4':[0,1,1,1,1]})我可以這樣創建一個列：df['added'] = df['v1'] + df['v2']*2 + df['v3']*4 + df['v4']*8我的想法是迭代這樣的變量列表（應該注意，在我的真正問題上，我不知道列數）：variables = ['v1', 'v2', 'v3', 'v4']df['added'] = df['v1']for ind, var in enumerate(variables[1:]) : df['added'] = df['added'] + df[var] << ind但是，這會引發錯誤：“TypeError：<<：'Series' 和 'int' 不支持的操作數類型。我可以用 pandas.DataFrame.apply() 解決我的問題：variables = ['v1', 'v2', 'v3', 'v4']df['added'] = df['v1']for ind, var in enumerate(variables[1:]) : df['added'] = df['added'] + df[var].apply(lambda x : x << ind )但是， apply （通常）很慢。我怎樣才能更有效地做事？
3 回答
肥皂起泡泡
TA貢獻1829條經驗 獲得超6個贊
使用這個解決方案，只是簡化，因為排序已經交換：
df['new'] = df.values.dot(1 << np.arange(df.shape[-1]))
print (df)
v1 v2 v3 v4 new
0 0 0 0 0 0
1 1 0 1 1 13
2 0 0 1 1 12
3 0 1 0 1 10
4 1 1 1 1 15
1000行和 4 列的性能：
np.random.seed(2019)
N= 1000
df = pd.DataFrame(np.random.choice([0,1], size=(N, 4)))
df.columns = [f'v{x+1}' for x in df.columns]
In [60]: %%timeit
...: df['new'] = df.values.dot(1 << np.arange(df.shape[-1]))
113 μs ± 1.45 μs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
尤卡解決方案：
In [65]: %%timeit
...: variables = ['v1', 'v2', 'v3', 'v4']
...: df['added'] = df['v1']
...: for ind, var in enumerate(variables[1:]) :
...: df['added'] = df['added'] + [x<<ind for x in df[var]]
...:
1.82 ms ± 16.2 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
原解決方案：
In [66]: %%timeit
...: variables = ['v1', 'v2', 'v3', 'v4']
...: df['added'] = df['v1']
...: for ind, var in enumerate(variables[1:]) :
...: df['added'] = df['added'] + df[var].apply(lambda x : x << ind )
...:
3.14 ms ± 8.52 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
慕桂英3389331
TA貢獻2036條經驗 獲得超8個贊
在回答您關于更有效替代方案的問題時，我發現列表理解確實對您有所幫助：
variables = ['v1', 'v2', 'v3', 'v4']
df['added'] = df['v1']
for ind, var in enumerate(variables[1:]) :
%timeit df['added'] = df['added'] + [x<<ind for x in df[var]]
308 μs ± 22.9 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
322 μs ± 19 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
316 μs ± 10.5 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
所以 315 μs 與：
variables = ['v1', 'v2', 'v3', 'v4']
df['added'] = df['v1']
for ind, var in enumerate(variables[1:]) :
%timeit df['added'] = df['added'] + df[var].apply(lambda x : x << ind )
500 μs ± 38.2 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
503 μs ± 32.1 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
481 μs ± 32 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
作為免責聲明，我不同意總和的價值，但這是一個不同的話題:)
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