Question

我想使用Python处理某些文件，并具有以下脚本：

import pandas as pd
loc1 = r'D:\1103\DC431.txt'
loc2 = r'D:\1103\DC432.txt'
loc3 = r'D:\1103\DC433.txt'
loc4 = r'D:\1103\DC434.txt'
loc5 = r'D:\1103\DC435.txt'
loc6 = r'D:\1103\DC436.txt'
loc7 = r'D:\1103\DC437.txt'
# Start
df1 = pd.read_table(loc1, sep=('\t'), engine='python')
df2 = pd.read_table(loc2, sep=('\t'), engine='python')
df3 = pd.read_table(loc3, sep=('\t'), engine='python')
df4 = pd.read_table(loc4, sep=('\t'), engine='python')
df5 = pd.read_table(loc5, sep=('\t'), engine='python')
df6 = pd.read_table(loc6, sep=('\t'), engine='python')
df7 = pd.read_table(loc7, sep=('\t'), engine='python')

s1 = df1[df1['No (Int)'] == 4]
s2 = df2[df2['No (Int)'] == 4]
s3 = df3[df3['No (Int)'] == 4]
s4 = df4[df4['No (Int)'] == 4]
s5 = df5[df5['No (Int)'] == 4]
s6 = df6[df6['No (Int)'] == 4]
s7 = df7[df7['No (Int)'] == 4]
# End

# Other scripts

我试图使用for循环简化上述“开始”到“结束”之间的脚本，但是python无法识别：

for n in range (1,8):
    dfn = pd.read_table(locn, sep=('\t'), engine='python')
    sn = dfn[dfn['No (Int)'] == 4]

有人有好主意吗？

Answer 1

dfn和sn是变量名，python不知道您的意思是索引n。您可以改用列表。首先没有列表理解：

import pandas as pd

locs = []
dfs = []
ss = []

for i in range(1, 8):
    locs[i] = r'D:\1103\DC43{}.txt'.format(i)
    dfs[i] = pd.read_csv(locs[i], sep='\t')
    ss[i] = dfs[i][dfs[i]['No (Int)'] == 4

具有列表理解功能：

import pandas as pd

locs = [r'D:\1103\DC43{}.txt'.format(i) for i in range(1,8)]
dfs = [pd.read_csv(loc, sep='\t') for loc in locs]
ss = [df[df['No (Int)'] == 4] for df in dfs]

注意：

pandas.read_table已被pandas.read_csv取代。
C引擎可能就足够了，仅在需要时再添加python引擎。

Answer 2

怎么样：

dict_result = dict()
for id in range(431, 438):
    loc = r'D:\1103\DC{0}.txt'.format(id)
    df = pd.read_table(loc, sep=('\t'), engine='python')
    s = df[df['No (Int)'] == 4]
    dict_result[id] = (loc, df, s)

Answer 3

您可以使用eval获取变量Xn的值，并使用exec执行一条语句。这是一个简单的示例：

a1 = 1
a2 = 2
b1 = {'1':{'w':1}, '2':2}
b2 = {'1':{'w':3}, '2':4}
for i in range(1,3):
    print(eval('a%d' % i))
    print(eval('b%d' % i)[str(i)])
exec('hello%d = 1234' % 99)
print(hello99)

输出为：

1
{'w': 1}
2
4
1234

在Python3.4.9上测试

带for循环的Python递归

3 个答案: