jun*_*er- 11 python numpy pandas
我有以下代码,它获取pandas数据帧的一列中的值,并使它们成为新数据框的列.数据帧第一列中的值将成为新数据帧的索引.
从某种意义上说,我想将邻接列表转换为邻接矩阵.这是迄今为止的代码:
import pandas as pa
print "Original Data Frame"
# Create a dataframe
oldcols = {'col1':['a','a','b','b'], 'col2':['c','d','c','d'], 'col3':[1,2,3,4]}
a = pa.DataFrame(oldcols)
print a
# The columns of the new data frame will be the values in col2 of the original
newcols = list(set(oldcols['col2']))
rows = list(set(oldcols['col1']))
# Create the new data matrix
data = np.zeros((len(rows), len(newcols)))
# Iterate over each row and fill in the new matrix
for row in zip(a['col1'], a['col2'], a['col3']):
rowindex = rows.index(row[0])
colindex = newcols.index(row[1])
data[rowindex][colindex] = row[2]
newf = pa.DataFrame(data)
newf.columns = newcols
newf.index = rows
print "New data frame"
print newf
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这适用于此特定实例:
Original Data Frame
col1 col2 col3
0 a c 1
1 a d 2
2 b c 3
3 b d 4
New data frame
c d
a 1 2
b 3 4
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如果col3中的值不是数字,它将失败.我的问题是,有没有更优雅/更健壮的方式来做到这一点?
unu*_*tbu 15
这看起来像枢轴的工作:
import pandas as pd
oldcols = {'col1':['a','a','b','b'], 'col2':['c','d','c','d'], 'col3':[1,2,3,4]}
a = pd.DataFrame(oldcols)
newf = a.pivot(index='col1', columns='col2')
print(newf)
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产量
col3
col2 c d
col1
a 1 2
b 3 4
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如果您不想要MultiIndex列,可以删除col3使用:
newf.columns = newf.columns.droplevel(0)
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然后会屈服
col2 c d
col1
a 1 2
b 3 4
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