Question

我有一个包含＆gt; 1M行的DataFrame。我想选择某个列包含某个子字符串的所有行：

matching = df['col2'].str.contains('substr', case=True, regex=False)
rows = df[matching].col1.drop_duplicates()

但这种选择很慢，我想加快速度。我们说我只需要第一个 n 结果。有没有办法在获得 n 结果后停止matching？我试过了：

matching = df['col2'].str.contains('substr', case=True, regex=False).head(n)

和

matching = df['col2'].str.contains('substr', case=True, regex=False).sample(n)

但他们没有更快。第二个语句是布尔值，非常快。我怎样才能加快第一个陈述？

Answer 1

信不信由你.str访问者很慢。您可以使用具有更好性能的列表推导。

df = pd.DataFrame({'col2':np.random.choice(['substring','midstring','nostring','substrate'],100000)})

测试平等

all(df['col2'].str.contains('substr', case=True, regex=False) ==
    pd.Series(['substr' in i for i in df['col2']]))

输出：

True

时序：

%timeit df['col2'].str.contains('substr', case=True, regex=False)
10 loops, best of 3: 37.9 ms per loop

与

%timeit pd.Series(['substr' in i for i in df['col2']])
100 loops, best of 3: 19.1 ms per loop

Answer 2

你可以用：

来表达

matching = df['col2'].head(n).str.contains('substr', case=True, regex=False)
rows = df['col1'].head(n)[matching==True]

但是，此解决方案会在第一个n行中检索匹配结果，而不是第一个n匹配结果。

如果您确实需要第一个n匹配结果，则应使用：

rows =  df['col1'][df['col2'].str.contains("substr")==True].head(n)

但是这个选项当然要慢一些。

受到@ ScottBoston答案的启发，您可以使用以下方法获得完整更快的解决方案：

rows = df['col1'][pd.Series(['substr' in i for i in df['col2']])==True].head(n)

这比使用此选项显示整个结果更快但速度更快。使用此解决方案，您可以获得第一个n匹配结果。

使用以下测试代码，我们可以看到每个解决方案的速度和结果：

import pandas as pd
import time

n = 10
a = ["Result", "from", "first", "column", "for", "this", "matching", "test", "end"]
b = ["This", "is", "a", "test", "has substr", "also has substr", "end", "of", "test"]

col1 = a*1000000
col2 = b*1000000

df = pd.DataFrame({"col1":col1,"col2":col2})

# Original option
start_time = time.time()
matching = df['col2'].str.contains('substr', case=True, regex=False)
rows = df[matching].col1.drop_duplicates()
print("--- %s seconds ---" % (time.time() - start_time))

# Faster option
start_time = time.time()
matching_fast = df['col2'].head(n).str.contains('substr', case=True, regex=False)
rows_fast = df['col1'].head(n)[matching==True]
print("--- %s seconds for fast solution ---" % (time.time() - start_time))


# Other option
start_time = time.time()
rows_other =  df['col1'][df['col2'].str.contains("substr")==True].head(n)
print("--- %s seconds for other solution ---" % (time.time() - start_time))

# Complete option
start_time = time.time()
rows_complete = df['col1'][pd.Series(['substr' in i for i in df['col2']])==True].head(n)
print("--- %s seconds for complete solution ---" % (time.time() - start_time))

这将输出：

>>> 
--- 2.33899998665 seconds ---
--- 0.302999973297 seconds for fast solution ---
--- 4.56700015068 seconds for other solution ---
--- 1.61599993706 seconds for complete solution ---

结果系列将是：

>>> rows
4     for
5    this
Name: col1, dtype: object
>>> rows_fast
4     for
5    this
Name: col1, dtype: object
>>> rows_other
4      for
5     this
13     for
14    this
22     for
23    this
31     for
32    this
40     for
41    this
Name: col1, dtype: object
>>> rows_complete
4      for
5     this
13     for
14    this
22     for
23    this
31     for
32    this
40     for
41    this
Name: col1, dtype: object

pandas：如何限制str.contains的结果？

2 个答案: