根据python中的索引位置转换值

我认为你需要:

df2.replace(df.iloc[0,0], 0)

将熊猫作为PD导入

df1 = pd.DataFrame({
    'a': [1,2],
    'b': [3,4]
})

df2 = pd.DataFrame({
    'a': [1,1],
    'b': [1,1]
})

df2.replace(df.iloc[0,0],0)

   a  b
0  0  0
1  0  0

df1 = pd.concat([df1]*10000)
df2 = pd.concat([df2]*10000)

%timeit df2.replace(df.iloc[0,0],0, inplace=True)
%timeit df2[df2 == df1.iloc[0, 0]] = 0

341 µs ± 9.36 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
1.24 ms ± 39 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

df = df2.mask(df2 == df1.iloc[0,0], 0)

df2[df2 == df1.iloc[0, 0]] = 0

或者:

df = pd.DataFrame(np.where(df2 == df1.iloc[0,0], 0, df2),index=df2.index,columns=df2.columns)

print (df)
   Out 1  Out 2  Out 3  Out 4  Out 5
0   7000   8000      0      0      0

print (df2 == df1.iloc[0,0])
   Out 1  Out 2  Out 3  Out 4  Out 5
0  False  False   True   True   True

计时

np.random.seed(100)

df1 = pd.DataFrame({
    'a': [1,2],
    'b': [3,4]
})


df2 = pd.DataFrame(np.random.randint(10, size=(1000,1000)))


In [106]: %timeit df2.replace(df1.iloc[0,0],0, inplace=True)
2.99 ms ± 324 µs per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [107]: %timeit df2.mask(df2 == df1.iloc[0,0], 0)
22.8 ms ± 1.71 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [108]: %timeit df2[df2 == df1.iloc[0, 0]] = 0
19.6 ms ± 497 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [109]: %timeit df = pd.DataFrame(np.where(df2 == df1.iloc[0,0], 0, df2),index=df2.index,columns=df2.columns)
5.81 ms ± 91.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

np.random.seed(100)

df1 = pd.DataFrame({
    'a': [1,2],
    'b': [3,4]
})

df2 = pd.DataFrame(np.random.randint(5, size=(1000,10)))

In [116]: %timeit df2.replace(df1.iloc[0,0],0, inplace=True)
856 µs ± 12.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [117]: %timeit df2.mask(df2 == df1.iloc[0,0], 0)
1.23 ms ± 25.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [118]: %timeit df2[df2 == df1.iloc[0, 0]] = 0
1.21 ms ± 4.26 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [119]: %timeit df = pd.DataFrame(np.where(df2 == df1.iloc[0,0], 0, df2),index=df2.index,columns=df2.columns)
445 µs ± 13.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)