Python Pandas vs R. Transformation Code Conciseness

时间：2014-01-23 18:02:58

标签： r python-3.x pandas

我转换了这个R代码：

# Raw data
data <- data.frame(
    metalname=c('Al','Cd','Cr','Co','Cu','Au','Fe','Pb','Mo','Ni','Pt','Au','Ta','Ti','W','Zn'),
    radius=c(0.1431,0.1490,0.1249,0.1253,0.1278,0.1442,0.1241,0.1750,0.1363,0.1246,0.1387,0.1445,0.1430,0.1445,0.1371,0.1332),
    crystal=c('FCC','HCP','BCC','HCP','FCC','FCC','BCC','FCC','BCC','FCC','FCC','FCC','BCC','HCP','BCC','HCP'))

# Calc lattice parameters (nm)
data <- rbind(
    transform(subset(data, crystal=='BCC'), N=2, latticea=4*radius/sqrt(3), latticec=0),
    transform(subset(data, crystal=='FCC'), N=4, latticea=2*radius*sqrt(2), latticec=0),
    transform(subset(data, crystal=='HCP'), N=6, latticea=2*radius, latticec=4*radius*sqrt(2/3))
)

这个熊猫代码：

import pandas as pd
import numpy as np
import math
from pandas import DataFrame

# Raw data
data = DataFrame({
    'metalname': ['Al','Cd','Cr','Co','Cu','Au','Fe','Pb','Mo','Ni','Pt','Au','Ta','Ti','W','Zn'],
    'radius': [0.1431,0.1490,0.1249,0.1253,0.1278,0.1442,0.1241,0.1750,0.1363,0.1246,0.1387,0.1445,0.1430,0.1445,0.1371,0.1332],
    'crystal': ['FCC','HCP','BCC','HCP','FCC','FCC','BCC','FCC','BCC','FCC','FCC','FCC','BCC','HCP','BCC','HCP']
})

# Calc lattice parameters (nm)
databcc = data[data.crystal=='BCC']
databcc['N'] = 2
databcc['latticea'] = 4*databcc.radius/math.sqrt(3)
datafcc = data[data.crystal=='FCC']
datafcc['N'] = 4
datafcc['latticea'] = 2*datafcc.radius/math.sqrt(2)
datahcp = data[data.crystal=='HCP']
datahcp['N'] = 6
datahcp['latticea'] = 2*datahcp.radius
datahcp['latticec'] = 4*datahcp.radius*math.sqrt(2/3)
data = databcc.append(datafcc).append(datahcp)

代码有效，但有没有办法让Python版本更简洁？理想情况下，我可以在没有R代码等临时变量的情况下一步完成多列计算。这可能吗？

5 个答案:

答案 0 :(得分：4)

这是pandas 0.13中新的query / eval功能的用例

databcc = data.query('crystal == "BCC"')
sqrt3 = sqrt(3)
databcc.eval('latticea = 4 * radius / sqrt3')

# ...

目前无法在表达式字符串中调用函数，因此您必须定义局部变量并使用它。

答案 1 :(得分：3)

这将是非常快的全部矢量化

In [65]: data.join(
              concat([ 
                DataFrame(dict(N=2, latticea=4*data.loc[data.crystal=='BCC','radius']/np.sqrt(3))), 
                DataFrame(dict(N=4, latticea=2*data.loc[data.crystal=='FCC','radius']/np.sqrt(2))), 
                DataFrame(dict(N=6, latticea=2*data.loc[data.crystal=='HCP','radius'], 
                                    latticec=4*data.loc[data.crystal=='HCP','radius']/np.sqrt(2/3.0))) 
                    ]))
Out[65]: 
   crystal metalname  radius  N  latticea  latticec
0      FCC        Al  0.1431  4  0.202374       NaN
1      HCP        Cd  0.1490  6  0.298000  0.729948
2      BCC        Cr  0.1249  2  0.288444       NaN
3      HCP        Co  0.1253  6  0.250600  0.613842
4      FCC        Cu  0.1278  4  0.180736       NaN
5      FCC        Au  0.1442  4  0.203930       NaN
6      BCC        Fe  0.1241  2  0.286597       NaN
7      FCC        Pb  0.1750  4  0.247487       NaN
8      BCC        Mo  0.1363  2  0.314771       NaN
9      FCC        Ni  0.1246  4  0.176211       NaN
10     FCC        Pt  0.1387  4  0.196151       NaN
11     FCC        Au  0.1445  4  0.204354       NaN
12     BCC        Ta  0.1430  2  0.330244       NaN
13     HCP        Ti  0.1445  6  0.289000  0.707903
14     BCC         W  0.1371  2  0.316619       NaN
15     HCP        Zn  0.1332  6  0.266400  0.652544

[16 rows x 6 columns]

答案 2 :(得分：1)

这相当于原始问题代码看起来比原始R杰作更好：

import pdb
import pandas as pd
import numpy as np
import math
from pandas import DataFrame

# Raw data
data = DataFrame({
    'metalname': ['Al','Cd','Cr','Co','Cu','Au','Fe','Pb','Mo','Ni','Pt','Au','Ta','Ti','W','Zn'],
    'radius': [0.1431,0.1490,0.1249,0.1253,0.1278,0.1442,0.1241,0.1750,0.1363,0.1246,0.1387,0.1445,0.1430,0.1445,0.1371,0.1332],
    'crystal': ['FCC','HCP','BCC','HCP','FCC','FCC','BCC','FCC','BCC','FCC','FCC','FCC','BCC','HCP','BCC','HCP']
})

def calc_lattic_params(x):
    N = None
    l = None
    lc = None
    if x['crystal'] == 'BCC':
        N = 2
        l = 4 * x['radius'] / math.sqrt(3)
    elif x['crystal'] == 'FCC':
        N = 4
        l = 2*x['radius'] / math.sqrt(2)
    elif x['crystal'] == 'HCP':
        N = 6
        l = 2*x['radius']
        lc = 4*x['radius']*math.sqrt(2.0/3.0)

    return pd.Series({'N': N, 'latticea': l, 'latticec': lc})

data = pd.concat([data, data.apply(calc_lattic_params, axis = 1)], axis = 1)

答案 3 :(得分：1)

如果有人感兴趣的话，可以使用Incanter（基于Lisp）版本进行比较：

(use '(incanter core stats charts))

; Raw data
(def data (dataset [:metalname :radius :crystal] [
    ["Al" 0.1431 "FCC"]
    ["Cd" 0.1490 "HCP"]
    ["Cr" 0.1249 "BCC"]
    ["Co" 0.1253 "HCP"]
    ["Cu" 0.1278 "FCC"]
    ["Au" 0.1442 "FCC"]
    ["Fe" 0.1241 "BCC"]
    ["Pb" 0.1750 "FCC"]
    ["Mo" 0.1363 "BCC"]
    ["Ni" 0.1246 "FCC"]
    ["Pt" 0.1387 "FCC"]
    ["Au" 0.1445 "FCC"]
    ["Ta" 0.1430 "BCC"]
    ["Ti" 0.1445 "HCP"]
    ["W" 0.1371 "BCC"]
    ["Zn" 0.1332 "HCP"]
]))

; Calc lattice parameters (nm)
(conj-rows
    (add-derived-column :latticec [] (fn [] 0)
    (add-derived-column :latticea [:radius] (fn [r] (/ (* 4 r) (sqrt 3)))
    (add-derived-column :n [] (fn [] 2)
        ($where {:crystal "BCC"} data))))
    (add-derived-column :latticec [] (fn [] 0)
    (add-derived-column :latticea [:radius] (fn [r] (* 2 r (sqrt 2)))
    (add-derived-column :n [] (fn [] 4)
        ($where {:crystal "FCC"} data))))
    (add-derived-column :latticec [:radius] (fn [r] (* 4 r (sqrt (/ 2 3))))
    (add-derived-column :latticea [:radius] (fn [r] (* 2 r))
    (add-derived-column :n [] (fn [] 6)
        ($where {:crystal "HCP"} data)))))

答案 4 :(得分：1)

由于问题是代码简洁性（Python vs R）的比较，这里是使用R的{​​{1}}解决方案：

data.table

设置library(data.table) dt <- data.table(data, key="crystal") data_transformed_dt <- rbind(dt["BCC", .(metalname, radius, crystal, N=2, latticea=4*radius/sqrt(3), latticec=0)], dt['FCC', .(metalname, radius, crystal, N=4, latticea=2*radius*sqrt(2), latticec=0)], dt['HCP', .(metalname, radius, crystal, N=6, latticea=2*radius, latticec=4*radius*sqrt(2/3))]) 的优点是它为水晶列编制索引，就像数据库中的索引一样。如果数据集非常大，这将极大地节省搜索时间（key="crystal"，"BCC" ...）。

另一种方法是创建另一个键 data.table，如下所示：

"FCC"

然后我们可以在加入时更新原始# v1.9.5+, for new feature "on = ", See github project page require(data.table) key = data.table(crystal = c("BCC", "FCC", "HCP"), latticea = c(4/sqrt(3), 2*sqrt(2), 2), latticec=c(0,0,4*sqrt(2/3)), N = c(2,4,6)) ，如下所示：

data

这应该是非常高效和快速的内存，因为我们通过引用更新（并且没有实现整个连接）。 setDT(data)[key , c("latticea", "latticec", "N") := .(radius * latticea, radius * latticec, N), on = "crystal"] # metalname radius crystal latticea latticec N # 1: Al 0.1431 FCC 0.4047479 0.0000000 4 # 2: Cd 0.1490 HCP 0.2980000 0.4866320 6 # 3: Cr 0.1249 BCC 0.2884442 0.0000000 2 # 4: Co 0.1253 HCP 0.2506000 0.4092281 6 # 5: Cu 0.1278 FCC 0.3614730 0.0000000 4 # 6: Au 0.1442 FCC 0.4078592 0.0000000 4 # 7: Fe 0.1241 BCC 0.2865967 0.0000000 2 # 8: Pb 0.1750 FCC 0.4949747 0.0000000 4 # 9: Mo 0.1363 BCC 0.3147714 0.0000000 2 # 10: Ni 0.1246 FCC 0.3524220 0.0000000 4 # 11: Pt 0.1387 FCC 0.3923028 0.0000000 4 # 12: Au 0.1445 FCC 0.4087077 0.0000000 4 # 13: Ta 0.1430 BCC 0.3302444 0.0000000 2 # 14: Ti 0.1445 HCP 0.2890000 0.4719350 6 # 15: W 0.1371 BCC 0.3166189 0.0000000 2 # 16: Zn 0.1332 HCP 0.2664000 0.4350294 6 在该列上执行连接，并在on = "crystal"中找到与data中每一行对应的匹配行索引，在这些匹配的行上，我们同时更新/创建必要的列。

请注意，数据的原始顺序也会保留在结果中。

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