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Python:在 Pandas Series 上使用 lambda 函数,if..else

转载 作者:行者123 更新时间:2023-11-28 20:21:21 42 4
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我正在尝试对存储在 pandas 系列对象中的一系列值应用过滤器。如果满足标准,则期望的输出是值本身,否则为零。我只能完成一半的工作:

criterion = testdata.map(lambda x: x < 30 or x > 60) 
testdata[criterion] =
Date
2015-01-05 62.358615
2015-01-06 64.349507
2015-01-13 61.726110
2015-01-14 63.574864
2015-01-15 66.020421
2015-01-16 63.041819
2015-01-20 61.458298
Name: rsi, dtype: float64

哪里:

testdata =
Date
2014-12-29 45.821073
2014-12-30 48.946830
2014-12-31 57.737593
2015-01-02 55.424047
2015-01-05 62.358615
2015-01-06 64.349507
2015-01-07 59.452935
2015-01-08 51.182115
2015-01-09 55.044419
2015-01-12 59.365017
2015-01-13 61.726110
2015-01-14 63.574864
2015-01-15 66.020421
2015-01-16 63.041819
2015-01-20 61.458298
2015-01-21 54.432635
2015-01-22 46.985675
2015-01-23 49.740981
2015-01-26 45.102370
2015-01-27 49.800855
Name: rsi, dtype: float64

但下面给出了完全不同的结果。显然,我不了解产生此结果的背景情况:

criterion2 = testdata.map(lambda x: x if (x < 30 or x > 60) else 0) 
testdata[criterion2]

结果:

rsi
0.000000 NaN
0.000000 NaN
0.000000 NaN
0.000000 NaN
62.358615 NaN
64.349507 NaN
0.000000 NaN
0.000000 NaN
0.000000 NaN
0.000000 NaN
61.726110 NaN
63.574864 NaN
66.020421 NaN
63.041819 NaN
61.458298 NaN
0.000000 NaN
0.000000 NaN
0.000000 NaN
0.000000 NaN
0.000000 NaN
Name: rsi, dtype: float64

我正在寻找与第一个输出相同的格式,除了在不满足条件的地方使用零。请帮忙。

最佳答案

你的

testdata.map(lambda x: x if (x < 30 or x > 60) else 0)

已经返回你想要的。

关于Python:在 Pandas Series 上使用 lambda 函数,if..else,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28195028/

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