Question

我通过一个非常大的字典列表进行过滤。 kept是全局列表，其中包含大约9000个词典，所有词典都具有相同的键。我试图删除所有包含“M_P”字典的词典。值大于-4.5并且有超过一半的值因此我创建了一个仅用于其目的的函数。当我检查它们是否已在后来的功能中被删除时，仍然剩下~3000。任何人都可以告诉我为什么会发生这种情况，我能相信这些功能会做我告诉它的事情吗？

def removeMag():

    countMag = 0
    for mag in kept:
        if to_float(mag['M_P']) > -4.5:
            kept.remove(mag)
            countMag += 1
        else:
            continue

    print '\n'
    print ' Number of mags > -4.5 actually removed: '
    print countMag

def remove_anomalies():    
    count = 0
    count08 = 0
    count09 = 0
    count01 = 0
    countMag = 0
    countMagDim = 0
    #Want to remove Q* < 15 degrees
    for row in kept:
        #to_float(kept(row))
        #Q* greater than 15
        if to_float(row['Q*']) < 15.00:
            kept.remove(row)
        elif to_float(row['vel']) > 80.00:
            kept.remove(row)
        elif to_float(row['err']) >= 0.5*to_float(row['vel']):
            kept.remove(row)

        elif row['log_10_m'] == '?':
            kept.remove(row)
            #print row
            count+=1
        elif row['M_P'] == '?':
            kept.remove(row)
            countMag += 1
        elif to_float(row['M_P']) > -4.5:
            countMagDim += 1

就在这里，我正在检查它。 ^^^

        elif to_float(row['T_j']) < -50.00 or to_float(row['T_j'] >    50.00):
        kept.remove(row)
        count01 += 1

        #make sure beg height is above end height.
        elif to_float(row['H_beg']) < to_float(row['H_end']):
            kept.remove(row)
        #make sure zenith distance is not greater than 90
        elif to_float(row['eta_p']) > 90.00:
            kept.remove(row)
        #Remove extremities hyperbolic orbits    
        elif (to_float(row['e']) > 2.00 and to_float(row['e']) == 0.00 and to_float(row['a']) == 0.00 and to_float(row['incl']) == 0.00 and to_float(row['omega']) == 0.00 and to_float(row['anode']) == 0.00 and to_float(row['alp_g']) == 0.00 and to_float(row['del_g']) == 0.00 and to_float(row['lam_g']) == 0.00 and to_float(row['bet_g']) == 0.00):
            kept.remove(row)
            count08+=1
        elif to_float(row['q_per']) == 0.00:
            kept.remove(row)
            count09+=1
        elif to_float(row['q_aph']) == 0.00:
            kept.remove(row)
            count09+=1
        else: continue

    print 'Number of dicts with ? as mass value:'
    print count    

    print " Number removed with orbital elements condition: "
    print count08

    print "Number of per or aph equal to 0: "
    print count09

    print "Number of T_j anomalies: "
    print count01

    print "Number of Magnitudes removed from '?': "
    print countMag

以下输出类似于3000。

    print "Number of Magnitudes to be removed from too dim: "
    print countMagDim   
'''    
    print "\n"
    print "log mass values:"
    for row2 in kept:
        print row2['log_10_mass']
    print "\n"
'''

Answer 1

使用for循环进行迭代时，Python不会自动复制列表，而是直接迭代它。因此，当您删除元素时，循环不会考虑更改，并将跳过列表的元素。

示例：

>>> l = [1,2,3,4,5]
>>> for i in l: l.remove(i)
>>> l
[2, 4]

您可以使用list indice作为简写，在迭代之前复制列表，例如：

>>> for i in l[:]: l.remove(i)
>>> l
[]

Answer 2

正如其他人所说，你在迭代它时修改一个数组。

这个简单的单行将是

kept = [mag for mag in kept if to_float(mag['M_P']) <= -4.5]

其中只保留您感兴趣的所有条目，替换原始列表。

计算删除的数量只是在理解之前和之后采用len(kept)并采取差异。

可替换地，

discarded = [mag for mag in kept if to_float(mag['M_P']) > -4.5]
kept = [mag for mag in kept if to_float(mag['M_P']) <= -4.5]

拆分数组而不会丢失任何信息

Answer 3

你永远不应该修改你在for循环中迭代的序列。只看你的第一个功能：

def removeMag():

    countMag = 0
    for mag in kept:
        if to_float(mag['M_P']) > -4.5:
            kept.remove(mag)
            countMag += 1

您正在循环中remove上呼叫kept。这会导致未指定的行为，任何事情都可能发生。请参阅this question。

解决此问题的一种简单方法是使用新列表来保留项目：

mag_to_keep = []
for mag in kept:
    if float(mag['M_P']) <= -4.5:
        mag_to_keep.append(mag)

kept = mag_to_keep

Python忽略命令从列表中删除项目

3 个答案: