遍历 pandas df 中的行

Question

我有如下所示的 df:

           CX     CY   CS
97539   0.39896 0.7787  0
97540   0.39896 0.7787  0
97541   0.39896 0.7787  0
97542   0.39896 0.7787  0
97543   0.39896 0.7787  0
97544   0.39896 0.7787  0
97545   0.39896 0.7787  0
97546   0.39896 0.7787  0
97547   0.39896 0.7787  0
97548   0.39896 0.7787  0
97549   0.39896 0.7787  0
97550   0.39896 0.7787  0
97551   0.39896 0.7787  0
97552   0.39896 0.7787  0
97553   0.39896 0.7787  0
97554   0.39896 0.7787  0
97555   0.39896 0.7787  0
97556   0.39896 0.7787  0
97557   0.39896 0.7787  0
97558   0.39896 0.7787  0
97559   0.39896 0.7787  0
97560   0.39896 0.7787  0
97561   0.39896 0.7787  1
97562   0.39896 0.7787  0
97563   0.39896 0.7787  0
97564   0.39896 0.7787  0
97565   0.39896 0.7787  0

我只想保留 df 的一部分，直到 'CS' 列上的值变为 1 并删除其余行。所以我想要这样的东西：

           CX     CY   CS
97539   0.39896 0.7787  0
97540   0.39896 0.7787  0
97541   0.39896 0.7787  0
97542   0.39896 0.7787  0
97543   0.39896 0.7787  0
97544   0.39896 0.7787  0
97545   0.39896 0.7787  0
97546   0.39896 0.7787  0
97547   0.39896 0.7787  0
97548   0.39896 0.7787  0
97549   0.39896 0.7787  0
97550   0.39896 0.7787  0
97551   0.39896 0.7787  0
97552   0.39896 0.7787  0
97553   0.39896 0.7787  0
97554   0.39896 0.7787  0
97555   0.39896 0.7787  0
97556   0.39896 0.7787  0
97557   0.39896 0.7787  0
97558   0.39896 0.7787  0
97559   0.39896 0.7787  0
97560   0.39896 0.7787  0
97561   0.39896 0.7787  1

有什么办法吗？请注意，1 的值可以在任何行，所以我不能只使用 .iloc()。理想情况下，我想避免 itterows().

Answer 1

如果总是有至少一个 1 是可能的比较值 Series.eq and then get index of first 1 by Series.idxmax, last filter by DataFrame.loc:

df1 = df.loc[: df['CS'].eq(1).idxmax()]

如果也没有 1 值，则解决方案有效 - 然后 return 空 DataFrame：

m = df['CS'].eq(1)
df1 = df.loc[: m.idxmax()] if m.any() else pd.DataFrame()

或使用Series.cummax in boolean indexing的技巧，只需要更改顺序2次：

df1 = df[df['CS'].iloc[::-1].eq(1).cummax().iloc[::-1]]

遍历 pandas df 中的行

Iterate over rows in pandas df

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