Python Pandas vs R.转换代码简洁

Question

我转换了此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))
)

对于此PANDAS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，但是有没有办法使Python版本更简洁？理想情况下，我将能够在一个步骤中进行多列计算，而无需像R代码这样的临时变量。这可能吗？

Answer 1

这是pandas中新的query/eval功能的用例0.13

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

当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。

Answer 2

这将非常快，因为它的所有矢量

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]

Answer 3

是否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)

Answer 4

如果有人有兴趣，这是用于比较的印加人（基于LISP）的版本：

(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)))))

Answer 5

由于问题是代码简洁（ Python vs R）的比较，这是使用data.table的R解决方案：

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))])

设置key="crystal"的优点在于，它像数据库中的索引一样索引了Crystal列。如果数据集非常大，这将节省搜索时间（"BCC"，"FCC" ...）。

---

另一种解决方法是创建另一个键 data.table，如下：

# 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))

然后，我们可以在加入时更新原始data：

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

这应该是非常有效且快速的，因为我们通过参考进行更新（并且没有整个联接未介绍）。on = "crystal"在该列上执行一个联接，并在data中找到与key中每一行相对应的匹配行索引，在那些匹配行中，我们同时更新/创建必要的列。

。

请注意，数据的原始顺序也保留在结果中。