查找大于级别的值 - Python Pandas
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Find value greater than level - Python Pandas
提问by Jared
In a time series (ordered tuples), what's the most efficient way to find the first time a criterion is met?
在时间序列（有序元组）中，找到第一次满足条件的最有效方法是什么？
In particular, what's the most efficient way to determine when a value goes over 100 for the value of a column in a pandas data frame?
特别是，确定 Pandas 数据框中列的值何时超过 100 的最有效方法是什么？
I was hoping for a clever vectorized solution, and not having to use df.iterrows().
我希望有一个聪明的矢量化解决方案，而不必使用df.iterrows().
For example, for price or count data, when a value exceeds 100. I.e. df['col'] > 100.
例如，对于价格或计数数据，当一个值超过 100 时。即 df['col'] > 100。
price
date
2005-01-01 98
2005-01-02 99
2005-01-03 100
2005-01-04 99
2005-01-05 98
2005-01-06 100
2005-01-07 100
2005-01-08 98
but for potentially very large series. Is it better to iterate (slow) or is there a vectorized solution?
但对于可能非常大的系列。迭代（慢）更好还是有矢量化解决方案？
A df.iterrows()solution could be:
一个df.iterrows()解决方案可能是：
for row, ind in df.iterrows():
if row['col'] > value_to_check:
breakpoint = row['value_to_record'].loc[ind]
return breakpoint
return None
But my question is more about efficiency (potentially, a vectorized solution that will scale well).
但我的问题更多是关于效率（可能是一个可以很好扩展的矢量化解决方案）。
回答by Merlin
Try this: "> 99"
试试这个："> 99"
df[df['price'].gt(99)].index[0]
returns "2", the second index row.
返回"2"，第二个索引行。
all row indexes greater than 99
所有大于 99 的行索引
df[df['price'].gt(99)].index
Int64Index([2, 5, 6], dtype='int64')
回答by user3304496
This will return the index value of the first occurrence of 100 in the series:
这将返回系列中第一次出现 100 的索引值：
index_value = (df['col'] - 100).apply(abs).idxmin()
If there is no value exactly 100, it should return the index of the closest value.
如果没有正好是 100 的值，它应该返回最接近值的索引。
