有没有一种简单的方法可以在Jupyter笔记本中快速查看两个pd.DataFrames的内容?
df1 = pd.DataFrame([(1,2),(3,4)], columns=['a', 'b'])
df2 = pd.DataFrame([(1.1,2.1),(3.1,4.1)], columns=['a', 'b'])
df1, df2
答案 0 :(得分:2)
你应该从@Wes_McKinney
尝试这个功能def side_by_side(*objs, **kwds):
''' Une fonction print objects side by side '''
from pandas.io.formats.printing import adjoin
space = kwds.get('space', 4)
reprs = [repr(obj).split('\n') for obj in objs]
print(adjoin(space, *reprs))
# building a test case of two DataFrame
import pandas as pd
import numpy as np
n, p = (10, 3) # dfs' shape
# dfs indexes and columns labels
index_rowA = [t[0]+str(t[1]) for t in zip(['rA']*n, range(n))]
index_colA = [t[0]+str(t[1]) for t in zip(['cA']*p, range(p))]
index_rowB = [t[0]+str(t[1]) for t in zip(['rB']*n, range(n))]
index_colB = [t[0]+str(t[1]) for t in zip(['cB']*p, range(p))]
# buliding the df A and B
dfA = pd.DataFrame(np.random.rand(n,p), index=index_rowA, columns=index_colA)
dfB = pd.DataFrame(np.random.rand(n,p), index=index_rowB, columns=index_colB)
side_by_side(dfA,dfB)
输出
cA0 cA1 cA2 cB0 cB1 cB2
rA0 0.708763 0.665374 0.718613 rB0 0.320085 0.677422 0.722697
rA1 0.120551 0.277301 0.646337 rB1 0.682488 0.273689 0.871989
rA2 0.372386 0.953481 0.934957 rB2 0.015203 0.525465 0.223897
rA3 0.456871 0.170596 0.501412 rB3 0.941295 0.901428 0.329489
rA4 0.049491 0.486030 0.365886 rB4 0.597779 0.201423 0.010794
rA5 0.277720 0.436428 0.533683 rB5 0.701220 0.261684 0.502301
rA6 0.391705 0.982510 0.561823 rB6 0.182609 0.140215 0.389426
rA7 0.827597 0.105354 0.180547 rB7 0.041009 0.936011 0.613592
rA8 0.224394 0.975854 0.089130 rB8 0.697824 0.887613 0.972838
rA9 0.433850 0.489714 0.339129 rB9 0.263112 0.355122 0.447154
答案 1 :(得分:0)
最接近你想要的是:
> df1.merge(df2, right_index=1, left_index=1, suffixes=("_1", "_2"))
a_1 b_1 a_2 b_2
0 1 2 1.1 2.1
1 3 4 3.1 4.1
这不是特定的笔记本电脑,但它会起作用,而且并不复杂。另一种解决方案是将您的数据帧转换为图像并将它们并排放置在子图中。但它有点牵强和复杂。
答案 2 :(得分:0)
我最终使用辅助函数快速比较两个数据帧:
def cmp(df1, df2, topn=10):
n = topn
a = df1.reset_index().head(n=n)
b = df2.reset_index().head(n=n)
span = pd.DataFrame(data=[('-',) for _ in range(n)], columns=['sep'])
a = a.merge(span, right_index=1, left_index=1)
return a.merge(b, right_index=1, left_index=1, suffixes=['_L', '_R'])