打开包含3列的txt文件，条件是检查第3列的差异是否为5，并将该行的第1列和第2列返回给新的txt fi

Question

我对编码和python非常陌生，我认为我正在咬下更多我能咀嚼的东西，但我正在尝试创建一个程序，读取一个包含3列信息的txt文件，然后获取这些列并列出它们。 然后，我想创建一个条件，将第3列的行值与自身上方和下方的行进行比较，如果值的差异大于5，它将复制第1列和第2列中的行，在第3列中找到该值，并将其附加到一个名为spikes的新列表中，我希望可以用它创建一个新的单独的txt文件。 名为“xyz”的txt文件值示例_测试.txt“：”

98015.985   -4922343.462    101.098 
98015.985   -4922343.712    101.098 
98015.985   -4922343.962    101.093 
98015.985   -4922344.212    101.089 
98015.985   -4922344.462    108.09 
98015.985   -4922344.712    101.095 
98015.985   -4922344.962    101.093 
98015.985   -4922345.212    101.083 
98015.985   -4922345.462    101.081 

到目前为止，我能得到并弄清楚的是：

 import csv,math listxy = [] listz = [] spikes = [] files =
 list(csv.reader(open('xyz_test.txt', 'rb'), delimiter='\t'))

 for z in files:
     listxy = z[0],z[1]
     listz = z[2]
     print listz

我得到的结果如下：

101.098 
101.098 
101.093 
101.089 
108.09 
101.095 
101.093 
101.083 
101.081

现在，我尝试运行一个条件，首先发现列表中的一个数字与其上下数字的差值大于5，但不断出现以下错误： “并非所有参数都在字符串格式化期间转换”
“无法连接'str'和'int'对象”

有谁能帮我一下吗。你知道吗

Thanks for all the help from everyone, learned allot. i have changed the code to fit what i need , here is what i ended up with. still tweaking , have to create something that sorts the values and loop through several txt files but this is what have so far:

from __future__ import print_function


import pandas as pd
# sets dipslay to larger extent
#pd.set_option('display.height', 10000000)
#pd.set_option('display.max_rows', 5000000)
#pd.set_option('display.max_columns', 50)
#pd.set_option('display.width', 10000)

limit = 3
tries = 0

while True:
        print ("----------------------------------------------------")
        spikewell = float(raw_input("Please Enter Parameters: "))
        tries += 1
        if tries == 4:
            print ("----------------------------------------------------")
            print ("Entered incorrectly to many times.....Exiting")
            print ("----------------------------------------------------")
            break
        else:
            if spikewell > 50:
               print ("parameters past limit (20)")
               print ("----------------------------------------------------")
               print (tries)
               continue
            elif spikewell < 0:
               print ("Parameters cant be negative")
               print ("----------------------------------------------------")
               print (tries)
               continue
            else:
               spikewell
               print ("Parameters are set")
               print (spikewell)
               print ("Searching files")
               print ("----------------------------------------------------")





        terrain = "1_tile_test.txt"
        for df in terrain:
            df = pd.read_csv('1_tile_test.txt', sep=r'\s+', names=['____x____  ','____y____  ','____z____'])
# print orginal data frame (for testing)

# get spikes's coordinates
# df['col3'].shift(1) - previous value of the 'col3' column
# df['col3'].shift(-1) - next value of the 'col3' column
            spikes = df.loc[(df['____z____'] - df['____z____'].shift(1) > spikewell) & \
            (df['____z____'] - df['____z____'].shift(-1) > spikewell)]
            wells = df.loc[-((df['____z____'] - df['____z____'].shift(1) > spikewell)) & \
            -((df['____z____'] - df['____z____'].shift(-1)) > -spikewell)]
# print and save spikes

   # print(spikes[['col1', 'col2','col3']])
   # print(spikes2[['col1', 'col2','col3']])
   # print(wells[['col1', 'col2','col3']])
   # print(wells2[['col1', 'col2','col3']])

            spikes[['____x____  ','____y____  ','____z____']].to_csv('spikes.txt', sep='\t', index=False)
            #spikes2[['____x____  ','____y____  ','____z____']].to_csv('spikes.txt', sep='\t', index=False)
            wells[['____x____  ','____y____  ','____z____']].to_csv('wells.txt', sep='\t', index=False)
            #wells2[['____x____  ','____y____  ','____z____']].to_csv('wells.txt', sep='\t', index=False)
            print ("----------------------------------------------------")
            print ('Search completed')
            break

        break

Answer 1

你可能想仔细看看pandas

输入数据（出于测试目的，我添加了一行[col3==111.110]）：

98015.985   -4922343.462    101.098 
98015.985   -4922343.712    101.098 
98015.985   -4922343.962    101.093 
98015.985   -4922344.212    101.089 
98015.985   -4922344.462    108.09 
98015.985   -4922344.712    101.095 
98015.985   -4922344.962    101.093 
98015.985   -4922345.212    101.083 
98015.985   -4922344.462    111.110 
98015.985   -4922345.462    101.081 

代码：

from __future__ import print_function

import pandas as pd

df = pd.read_csv('data.csv', sep=r'\s+', names=['col1','col2','col3'])
# print orginal data frame (for testing)
print(df)

# get spikes's coordinates
# df['col3'].shift(1) - previous value of the 'col3' column
# df['col3'].shift(-1) - next value of the 'col3' column
spikes = df.loc[(df['col3'] - df['col3'].shift(1) > 5) & (df['col3'] - df['col3'].shift(-1) > 5)]

# print and save spikes
print(spikes[['col1', 'col2']])
spikes[['col1', 'col2']].to_csv('spikes.csv', sep='\t', index=False)

输出：

        col1         col2     col3
0  98015.985 -4922343.462  101.098
1  98015.985 -4922343.712  101.098
2  98015.985 -4922343.962  101.093
3  98015.985 -4922344.212  101.089
4  98015.985 -4922344.462  108.090
5  98015.985 -4922344.712  101.095
6  98015.985 -4922344.962  101.093
7  98015.985 -4922345.212  101.083
8  98015.985 -4922344.462  111.110
9  98015.985 -4922345.462  101.081
        col1         col2
4  98015.985 -4922344.462
8  98015.985 -4922344.462

你知道吗尖峰.csv地址：

col1    col2
98015.985   -4922344.462
98015.985   -4922344.462

Answer 2

举个例子：

import csv

def is_spike(three):
    first, second, third = three
    return abs(float(first[2]) - float(second[2])) > 5 and abs(float(second[2]) - float(third[2])) > 5

with open("yourcsvfile.csv") as csvfile:
    reader = csv.reader(csvfile)
    rows = list(reader)
    threes = zip(rows, rows[1:], rows[2:])
    spikes = [three for three in threes if is_spike(three)]

print spikes

输出（中间行为“峰值”）：

[(['98015.985', '-4922344.212', '101.089'], ['98015.985', '-4922344.462', '108.09'], ['98015.985', '-4922344.712', '101.095'])]

演练：

首先，我们使用csv模块读取整行数据，该模块为我们拆分数据。确保正确设置分隔符。您也可以手动读取它们，但这是更通用的。你知道吗

其次，我们压缩所有threes（如三行），并使用is_spike函数检查它们是否形成一个“尖峰”，这个函数非常简单。你知道吗

祝你好运。你知道吗