嗨,我有一个从两个文件中读取数据的函数。我想要做的是,外部循环开始读取第一个文件,如果第一个文件中的lum大于定义值(LC),则循环跳到下一个迭代。如果不是,代码会移动到内部循环,如果idc和id是相同的,则内部循环读取hlist并生成x、y、z列表。如果idc和id不是相同的值,则跳到内部循环的下一次迭代。当我打印测试状态时,似乎内部循环没有迭代,我不清楚为什么。我很感谢你的帮助
代码
def read_file(F): #Function that reads data froma file #and extracts specific data columns
X_pos = []
Y_pos = [] # Creats Data Lists
Z_pos = []
Idh = []
Id = []
LC = float(sys.argv[1])
N = 11#912639 # number of lines to be read
Nl = 11#896030
fl = open(Fl) #opens catalog file
fl.readline()
nlines_catalog = islice(fl, Nl)
f = open(F) #Opens hlist file
f.readline() # Strips Header
nlines_hlist = islice(f, N) #slices file to only read N lines
for linel in nlines_catalog:
if linel != '':
linel = linel.strip()
linel = linel.replace('\t', '')
columnsl = linel.split()
lum = float(columnsl[1])
id_catalog = int(columnsl[0])
if lum >= LC:
continue
print("lum1 =", lum)
#Id.append(idc)
print("id_catalog=", id_catalog)
for line in nlines_hlist:
if line != '':
line = line.strip()
line = line.replace('\t', ' ')
columns = line.split()
id_hlist = int(columns[1])
#Idh.append(id)
if id_hlist != id_ccatalog:
continue
print('idc =', idc, 'id =', id)
x = columns[17]
y = columns[18]
z = columns[19]
X_pos.append(x)
Y_pos.append(y) #appends data in list
Z_pos.append(z)
print(X)
X = [float(p) for p in X_pos]
Y = [float(p) for p in Y_pos]
Z = [float(p) for p in Z_pos]
Xa = numpy.array(X, dtype=float)
Ya = numpy.array(Y, dtype=float)
Za = numpy.array(Z, dtype=float)
return(Xa, Ya, Za)编辑
更改内环,使其重新设置,现在工作。
if id_catalog == id_halo:
print('id_catalog =',id_catalog,'id_halo =',id_halo)
x = columns[17] # assigns variable to columns
y = columns[18]
z = columns[19]
#vx = columns[]
#vy = columns[]
#vz = columns[]
X_pos.append(x)
Y_pos.append(y) #appends data in list
Z_pos.append(z)
break编辑I无法复制打印语句中的原始输出,并进行了编辑以反映这一事实
VirtualBox:~$ python /home/Astrophysics/Count_FixedLoop.py -21.5 125
('lum1 =', -21.78545)
('idc=', 2701276876L)
('idc =', 2701276876L, 'id =', 2701276876L)
('lum1 =', -21.69835)
('idc=', 2699751347L)
('lum1 =', -21.69942)
('idc=', 2699724518L)
('lum1 =', -21.74543)
('idc=', 2699724331L)
('lum1 =', -21.60912)
('idc=', 2699724726L)
('lum1 =', -21.53862)
('idc=', 2699725014L)
('lum1 =', -21.53155)
('idc=', 2701277269L)
['34.57223']这就是我期望的输出。
('lum1 =', -21.78545)
('idc=', 2701276876L)
('idc =', 2701276876L, 'id =', 2701276876L)
('lum1 =', -21.69835)
('idc=', 2699751347L)
('idc =', 2699751347L, 'id =', 2699751347L)
('lum1 =', -21.69942)
('idc=', 2699724518L)
('idc =', 2699724518L, 'id =', 2699724518L)
('lum1 =', -21.74543)
('idc=', 2699724331L)
('idc =', 2699724331L, 'id =', 2699724331L)
('lum1 =', -21.60912)
('idc=', 2699724726L)
('idc =', 2699724726L, 'id =', 2699724726L)
('lum1 =', -21.53862)
('idc=', 2699725014L)
('idc =', 2699725014L, 'id =', 2699725014L)
('lum1 =', -21.53155)
('idc=', 2701277269L)编辑
Fl文件的几行示例,其中idc数字为列,lum值为column1。我在第一个循环中用粗体打印了满足条件的idc值。
Format: ID, scatter = 0 0.05 0.1 0.13 0.15 0.16 0.18 0.2 0.25 0.3
**2701276876 -21.78545** -21.73791 -21.68872 -21.11125 -20.88102 -22.04709 -21.41715 -20.56944 -20.36757 -19.69895
**2699751347 -21.69835** -21.67935 -21.92425 -21.03465 -21.56561 -21.42124 -21.72893 -20.78131 -20.76342 -20.34830
**2699724518 -21.69942** -21.58352 -21.71149 -21.16240 -21.18507 -22.00277 -21.81500 -20.36141 -20.78227 -20.65697编辑
F文件的示例行,其中包含与第一个文件匹配的id和相应位置
#Scale(0) Id(1) Desc_scale(2) Descid(3) Num_prog(4) Pid(5) Upid(6) Desc_pid(7) Phantom(8) Mvir(9) Orig_Mvir(10) Rvir(11) Rs(12) Vrms(13) Mmp(14) Last_mm(15) Vmax(16) X(17) Y(18) Z(19)
0.9523 **2701276876** 0.9583 2714557311 1 -1 -1 -1 0 3.56533e+13 3.56100e+13 695.459000 80.562000 548.820000 1 0.3603 561.490000 **34.57223 140.20813 130.81985** -110.000 323.430 -123.520 3.56533e+13 3.56533e+13 561.490000 599.410000 7.539e+14 -3.799e+12 -1.992e+14 0.10259
0.9523 **2699751347** 0.9583 2713034575 4 -1 -1 -1 0 3.36604e+13 3.31300e+13 678.981000 111.199000 500.400000 1 0.8083 514.010000 **28.70439 138.70247 138.52176** -215.310 252.520 -120.970 3.36604e+13 3.36604e+13 514.010000 599.250000 5.516e+14 1.044e+14 6.133e+14 0.10973
0.9523 **2699724518** 0.9583 2713007786 1 -1 -1 -1 0 2.98000e+13 2.97500e+13 654.997000 87.324000 457.460000 1 0.4863 514.660000 **8.01627 135.31783 123.13322** -178.990 558.900 1.250 2.98000e+13 2.98000e+13 514.660000 514.660000 8.529e+14 2.711e+14 -3.624e+14 0.15137我期望的是,当两个文件中的id匹配时,第二个循环将追加X、Y和Z位置列表。因此,在本例中打印列表将给出
X = [34.57223,28.70439,8.01627]
Y = [140.20813,138.70247,135.31783]
Z = [130.81985,138.52176,123.13322]发布于 2012-07-24 19:15:25
您似乎希望nlines表现得像一个列表。然而,它却是一种http://docs.python.org/library/itertools.html,正如Ignacio上面指出的,它将被消耗一次。换句话说,内环不会在随后的外部循环执行时将其“重置”到第一行/索引。
考虑一下下面的模拟(我认为)你正在做的事情。以下是两个数据文件:
Data1:
file 1: one
file 1: two
file 1: three
file 1: four
file 1: fiveData2:
file 2: one
file 2: two
file 2: three
file 2: four
file 2: five
file 2: six运行此命令:
from itertools import islice
f1 = open ("Data1")
f2 = open ("Data2")
iterator1 = islice (f1, 3)
iterator2 = islice (f2, 3)
for line1 in iterator1:
print line1
for line2 in iterator2:
print line2在以下方面的成果:
file 1: one
file 2: one
file 2: two
file 2: three
file 1: two
file 1: three虽然人们可能错误地认为,data2的3行内容将为data1的前3行中的每一行打印。
因此,内部循环的第一次执行完全消耗了iterator2。在您自己的代码中,当id == idc时没有内部循环中断-换句话说,您在内环第一次执行时完全使用了interator nlines。
另一个例子请参见Python: itertools.islice not working in a loop。
一种解决方案可能是在id == idc时中断内部循环,但这将假设(我认为)在第二个文件中排序索引。您可以考虑为内部循环实际使用一个列表,尽管考虑到实际(非测试)数据的大小,这似乎是内存密集型的。很明显,您可以重新阅读第二个文件,尽管性能会受到影响。
https://stackoverflow.com/questions/11605346
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