[英]Python Pandas - Randomly select rows from a dataframe between 2 two rows
[英]How to select rows in a DataFrame between two values, in Python Pandas?
我正在尝试修改 DataFrame df
以仅包含列closing_price
中的值在 99 到 101 之间的行,并尝试使用以下代码执行此操作。
但是,我收到错误
ValueError:系列的真值不明确。 使用 a.empty、a.bool()、a.item()、a.any() 或 a.all()
我想知道是否有办法在不使用循环的情况下做到这一点。
df = df[(99 <= df['closing_price'] <= 101)]
还要考虑以下之间的系列:
df = df[df['closing_price'].between(99, 101)]
您应该使用()
对布尔向量进行分组以消除歧义。
df = df[(df['closing_price'] >= 99) & (df['closing_price'] <= 101)]
有一个更好的选择 - 使用query()方法:
In [58]: df = pd.DataFrame({'closing_price': np.random.randint(95, 105, 10)})
In [59]: df
Out[59]:
closing_price
0 104
1 99
2 98
3 95
4 103
5 101
6 101
7 99
8 95
9 96
In [60]: df.query('99 <= closing_price <= 101')
Out[60]:
closing_price
1 99
5 101
6 101
7 99
更新:回答评论:
我喜欢这里的语法,但在尝试与表达结合时失败了;
df.query('(mean + 2 *sd) <= closing_price <=(mean + 2 *sd)')
In [161]: qry = "(closing_price.mean() - 2*closing_price.std())" +\
...: " <= closing_price <= " + \
...: "(closing_price.mean() + 2*closing_price.std())"
...:
In [162]: df.query(qry)
Out[162]:
closing_price
0 97
1 101
2 97
3 95
4 100
5 99
6 100
7 101
8 99
9 95
newdf = df.query('closing_price.mean() <= closing_price <= closing_price.std()')
或者
mean = closing_price.mean()
std = closing_price.std()
newdf = df.query('@mean <= closing_price <= @std')
如果您正在处理多个值和多个输入,您还可以设置这样的应用函数。 在这种情况下,过滤位于特定范围内的 GPS 位置的数据帧。
def filter_values(lat,lon):
if abs(lat - 33.77) < .01 and abs(lon - -118.16) < .01:
return True
elif abs(lat - 37.79) < .01 and abs(lon - -122.39) < .01:
return True
else:
return False
df = df[df.apply(lambda x: filter_values(x['lat'],x['lon']),axis=1)]
如果必须重复调用pd.Series.between(l,r)
(对于不同的边界l
和r
),则会不必要地重复大量工作。 在这种情况下,对帧/系列进行一次排序然后使用pd.Series.searchsorted()
是有益的。 我测量了高达 25 倍的加速,见下文。
def between_indices(x, lower, upper, inclusive=True):
"""
Returns smallest and largest index i for which holds
lower <= x[i] <= upper, under the assumption that x is sorted.
"""
i = x.searchsorted(lower, side="left" if inclusive else "right")
j = x.searchsorted(upper, side="right" if inclusive else "left")
return i, j
# Sort x once before repeated calls of between()
x = x.sort_values().reset_index(drop=True)
# x = x.sort_values(ignore_index=True) # for pandas>=1.0
ret1 = between_indices(x, lower=0.1, upper=0.9)
ret2 = between_indices(x, lower=0.2, upper=0.8)
ret3 = ...
基准
测量pd.Series.between()
重复评估 ( n_reps=100
) 以及基于pd.Series.searchsorted()
的方法,对于不同的参数lower
和upper
。 在使用 Python v3.8.0 和 Pandas v1.0.3 的 MacBook Pro 2015 上,以下代码产生以下输出
# pd.Series.searchsorted()
# 5.87 ms ± 321 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# pd.Series.between(lower, upper)
# 155 ms ± 6.08 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
# Logical expressions: (x>=lower) & (x<=upper)
# 153 ms ± 3.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
import numpy as np
import pandas as pd
def between_indices(x, lower, upper, inclusive=True):
# Assumption: x is sorted.
i = x.searchsorted(lower, side="left" if inclusive else "right")
j = x.searchsorted(upper, side="right" if inclusive else "left")
return i, j
def between_fast(x, lower, upper, inclusive=True):
"""
Equivalent to pd.Series.between() under the assumption that x is sorted.
"""
i, j = between_indices(x, lower, upper, inclusive)
if True:
return x.iloc[i:j]
else:
# Mask creation is slow.
mask = np.zeros_like(x, dtype=bool)
mask[i:j] = True
mask = pd.Series(mask, index=x.index)
return x[mask]
def between(x, lower, upper, inclusive=True):
mask = x.between(lower, upper, inclusive=inclusive)
return x[mask]
def between_expr(x, lower, upper, inclusive=True):
if inclusive:
mask = (x>=lower) & (x<=upper)
else:
mask = (x>lower) & (x<upper)
return x[mask]
def benchmark(func, x, lowers, uppers):
for l,u in zip(lowers, uppers):
func(x,lower=l,upper=u)
n_samples = 1000
n_reps = 100
x = pd.Series(np.random.randn(n_samples))
# Sort the Series.
# For pandas>=1.0:
# x = x.sort_values(ignore_index=True)
x = x.sort_values().reset_index(drop=True)
# Assert equivalence of different methods.
assert(between_fast(x, 0, 1, True ).equals(between(x, 0, 1, True)))
assert(between_expr(x, 0, 1, True ).equals(between(x, 0, 1, True)))
assert(between_fast(x, 0, 1, False).equals(between(x, 0, 1, False)))
assert(between_expr(x, 0, 1, False).equals(between(x, 0, 1, False)))
# Benchmark repeated evaluations of between().
uppers = np.linspace(0, 3, n_reps)
lowers = -uppers
%timeit benchmark(between_fast, x, lowers, uppers)
%timeit benchmark(between, x, lowers, uppers)
%timeit benchmark(between_expr, x, lowers, uppers)
而不是这个
df = df[(99 <= df['closing_price'] <= 101)]
你应该用这个
df = df[(df['closing_price']>=99 ) & (df['closing_price']<=101)]
我们必须使用 NumPy 的按位逻辑运算符 |、&、~、^ 进行复合查询。 此外,括号对于运算符的优先级很重要。
有关更多信息,您可以访问链接: 比较、掩码和布尔逻辑
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