如何在 Python Pandas 中的两个值之间选择 DataFrame 中的行？

Question

我正在尝试将 DataFrame df 修改为仅包含 closing_price 列中的值介于 99 和 101 之间的行，并尝试使用下面的代码执行此操作。

但是，我得到了错误

ValueError：Series 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)]

Answer 1

您应该使用() 对您的布尔向量进行分组以消除歧义。

df = df[(df['closing_price'] >= 99) & (df['closing_price'] <= 101)]

Answer 2

还有一个更好的选择——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：回复评论：

我喜欢这里的语法，但在尝试与 表达； 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

Answer 3

也可以考虑series between：

df = df[df['closing_price'].between(99, 101)]

Answer 4

newdf = df.query('closing_price.mean() <= closing_price <= closing_price.std()')

或

mean = closing_price.mean()
std = closing_price.std()

newdf = df.query('@mean <= closing_price <= @std')

Answer 5

而不是这个

df = df[(99 <= df['closing_price'] <= 101)]

你应该使用这个

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 |、&、~、^ 来进行复合查询。 此外，括号对于运算符优先级很重要。

更多信息，您可以访问链接 :Comparisons, Masks, and Boolean Logic

Answer 6

你也可以使用.between()方法

emp = pd.read_csv("C:\\py\\programs\\pandas_2\\pandas\\employees.csv")

emp[emp["Salary"].between(60000, 61000)]

输出

Answer 7

如果您要处理多个值和多个输入，您还可以像这样设置一个应用函数。在这种情况下，为特定范围内的 GPS 位置过滤数据帧。

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)]

Answer 8

如果必须调用pd.Series.between(l,r) 重复（对于不同的界限l 和r），很多工作都是不必要的重复。在这种情况下，最好对帧/系列进行一次排序，然后使用pd.Series.searchsorted()。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 = ...

---

基准测试

针对lower和upper的不同参数测量pd.Series.between()的重复评估（n_reps=100）以及基于pd.Series.searchsorted()的方法。在我的带有 Python v3.8.0 和 Pandas v1.0.3 的 MacBook Pro 2015 上，下面的代码会产生以下输出

# 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)