逐行比较两个熊猫数据帧

Zubo 发表于 Dev

祖波

我有2个数据框，df1和df2，并要执行以下操作，将结果存储在中df3：

for each row in df1:

    for each row in df2:

        create a new row in df3 (called "df1-1, df2-1" or whatever) to store results 

        for each cell(column) in df1: 

            for the cell in df2 whose column name is the same as for the cell in df1:

                compare the cells (using some comparing function func(a,b) ) and, 
                depending on the result of the comparison, write result into the 
                appropriate column of the "df1-1, df2-1" row of df3)

例如，类似：

df1
A   B    C      D
foo bar  foobar 7
gee whiz herp   10

df2
A   B   C      D
zoo car foobar 8

df3
df1-df2 A             B              C                   D
foo-zoo func(foo,zoo) func(bar,car)  func(foobar,foobar) func(7,8)
gee-zoo func(gee,zoo) func(whiz,car) func(herp,foobar)   func(10,8)

我从这里开始：

for r1 in df1.iterrows():
    for r2 in df2.iterrows():
        for c1 in r1:
            for c2 in r2:

但不确定该如何处理，将不胜感激。

星狐

因此，要继续在评论中进行讨论，可以使用矢量化，它是pandas或numpy之类的图书馆的卖点之一。理想情况下，您永远不应该打电话iterrows()。我的建议要更明确一点：

# with df1 and df2 provided as above, an example
df3 = df1['A'] * 3 + df2['A']

# recall that df2 only has the one row so pandas will broadcast a NaN there
df3
0    foofoofoozoo
1             NaN
Name: A, dtype: object

# more generally

# we know that df1 and df2 share column names, so we can initialize df3 with those names
df3 = pd.DataFrame(columns=df1.columns) 
for colName in df1:
    df3[colName] = func(df1[colName], df2[colName])

现在，您甚至可以通过创建lambda函数，然后使用列名称压缩它们，将不同的函数应用于不同的列：

# some example functions
colAFunc = lambda x, y: x + y
colBFunc = lambda x, y; x - y
....
columnFunctions = [colAFunc, colBFunc, ...]

# initialize df3 as above
df3 = pd.DataFrame(columns=df1.columns)
for func, colName in zip(columnFunctions, df1.columns):
    df3[colName] = func(df1[colName], df2[colName])

想到的唯一“陷阱”是，您需要确保您的函数适用于列中的数据。例如，如果您要执行类似的操作df1['A'] - df2['A']（使用提供的df1，df2），则将产生a，ValueError因为未定义两个字符串的减法。只是要注意的事情。

编辑，重新：您的评论：这也是可行的。遍历更大的dfX.columns，因此您不会碰到KeyError，并if在其中抛出一条语句：

# all the other jazz
# let's say df1 is [['A', 'B', 'C']] and df2 is [['A', 'B', 'C', 'D']]
# so iterate over df2 columns
for colName in df2:
    if colName not in df1:
        df3[colName] = np.nan # be sure to import numpy as np
    else:
        df3[colName] = func(df1[colName], df2[colName])

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